1.1◉ It is clear from the fourth book that the forms of visible objects are reflected from polished bodies, and sight apprehends them in polished bodies according to reflection. And it has been shown how objects are apprehended through the reflection of [their] forms. Moreover, sight perceives the visible object at a determinate and principal location of reflection when there is no change in
the spatial disposition of the visible object with respect to the eye. The form perceived in a polished body is called an image. And in this book we shall explain image-locations in polished bodies, and we shall discuss how to gain a scientific understanding of these locations, as well as how to establish that science by demonstration, and [we shall also discuss] how [those] image-locations are rationally determined.

2.1◉ The image-location of any point is the point where the line of reflection intersects the normal imagined [to extend] from a point on a visible object to the line tangent to the common section of the surface of the mirror and the plane of reflection, or [to the common section] of the plane that coincides with [the plane of the] mirror and the plane of reflection. We shall demonstrate
[this point as follows].⁑

2.2◉ Take a plane mirror, set it up horizontally, and stand a smooth, straight wooden rod upright upon the mirror. Let the mirror be large enough that the whole rod can be seen [in it], for, unless the whole rod is visible, error can arise. Let a black spot be marked on the rod. To the eye a rod the same size as this one will appear behind the mirror and directly in line with the actual rod
and orthogonal to the mirror, and in the rod that is seen [in the mirror], the black spot marked [on the actual rod] will appear to lie the same distance behind the mirror’s surface as [it lies] on the rod above that surface. Furthermore, if the [actual] rod is inclined to the mirror, the rod seen [in the mirror] will appear inclined with the same slant, and the spot marked on the rod seen [in the mirror] will appear to lie the same distance from the mirror’s
surface as the spot marked [on the actual rod]. Moreover, if some [other] rod is dropped orthogonally to the mirror from the spot marked [on the actual rod], this rod will also appear [behind and] orthogonal to the mirror from the point on the apparent rod, and [it will appear] right in line with the actual rod. The same thing happens for multiple points marked on the rod. The very same thing will happen if the mirror is raised or lowered [by tilting].

2.3◉ From this, then, it is evident that the image of a visible point appears on the normal extended from the visible point to the surface of the mirror, and that in this mirror what is perpendicular to the mirror’s surface is perpendicular to the common section of the mirror’s surface and the [plane of] reflection.

2.4◉ The same thing can be shown in the case of a right cone placed upright with its base lying flat upon the plane mirror, for [in that case] another cone will appear [behind the mirror] directly in line with it and sharing the same base, the vertices of these cones lying the same distance from the [surface of the] mirror. And it is obvious that, if a straight line is extended from vertex
to vertex, it will be perpendicular to the [shared] base, and thus to the [surface of the] mirror, since that surface and the base [of the cones] coincide, so the [image of] the cone’s vertex will appear on the normal dropped from that vertex to the mirror. Likewise, from some point on the cone draw a line to the corresponding point on the cone seen [behind the mirror]. That line will be orthogonal to both the [cone’s] base and the mirror’s surface, so the image
of any point on the cone lies on the perpendicular imagined [to extend] from that point to the mirror’s surface.⁑

2.5◉ But whatever point on a body may face a plane mirror, a cone can be imagined with its vertex at that point, and this cone is orthogonal to its base as well as to the surface of the mirror or its continuation. And another cone opposite this one can be imagined sharing the same base and orthogonal to the mirror, and the perpendicular [extended] from vertex to vertex [of the two cones]
will be orthogonal to the mirror, so the image of any point facing the mirror lies on the normal [extended] from the point to the surface of the mirror or to its extension. But it is obvious that in [all] mirrors the perception of forms will be attained only along lines of reflection, so the image of the point seen [in the mirror] lies on a line of reflection, and every such line is straight, so the image of any point [seen in a mirror] lies at the
intersection-point of the normal [dropped] from that point to the surface of the mirror and the line of reflection. And it falls on plane mirrors, and the common section of the mirror’s surface and the plane of reflection is unique [in coinciding] with the line tangent to the point of reflection, so it is evident that in plane mirrors the appropriate image-location is the point where the normal [dropped] from the visible point to the line tangent to the common
section of the mirror’s surface and the plane of reflection intersects the line of reflection.⁑

2.6◉ In the case of spherical mirrors that are polished on the outer surface [i.e., convex] what we have claimed will be evident. Find [such] a mirror with a surface large enough that the [entire] form of the thin rod [just used with the plane mirror] can be seen when the rod is stood upright on it. The form of the rod [seen in the mirror] will appear directly in line with the rod [itself],
and on the form of the rod [the form of] the point marked on it will appear [to lie] the same [relative] distance behind the mirror’s surface as it lies [above the mirror] on the actual rod.⁑
Moreover, if the rod is narrower at one end than at the other, its form will appear conical in this kind of mirror, but this involves a visual illusion that we will explain later.⁑

2.7◉ Now, along with this rod form a cone that is orthogonal to a circular base that is rounded off perfectly, and let it also be applied to this mirror. A cone continuous with this one will appear [behind the mirror and] standing on the same base, but [it will appear] smaller than the [actual] cone. That the image should appear conical is clear from the fact that all the lines [extended]
from the vertex of the cone seen [in the mirror] to the circle at the base appear equal, and if the cone is moved a bit on the mirror’s surface away from where all of it is visible, i.e., so that some of it disappears from view, then, as long as the point of reflection on the mirror is exposed to sight, the image of the cone will be visible.⁑ And if the
eye is drawn away from the mirror or approaches it, provided it follows the [original] line of reflection extending between it and the point of reflection, the image of the cone will be visible, but the movement [of the eye] toward or away will be along this line in such a way that the point of reflection is marked. Extend a line from the location of the eye to the mark according to which the movement occurs.

2.8◉ Now, since the image of the cone is orthogonal to the cone’s base, and since that base forms one of [an infinite number of possible] circles on the sphere, the line [extended] from the vertex of the [actual] cone to the vertex of its image will be orthogonal to that circle, and it will pass through its center. It will also be orthogonal to the sphere and will pass through the sphere’s
center, and it will be orthogonal to the plane tangent to the sphere at the point where this line passes through [its surface]. Likewise, it will be orthogonal to the line tangent to the [great] circle on the sphere that passes through that point, and this tangent forms the common section of the plane of reflection and the plane tangent to the sphere at that point, and this line is tangent to the circle on the sphere that forms the common section of the sphere’s
surface and the plane of reflection. Thus, the line extended from the vertex of the cone to the vertex of its image is perpendicular to the line tangent to the common section of the plane of reflection and the mirror’s surface, and that section is a circle.

2.9◉ Therefore, the image of the vertex will be seen on this perpendicular, and it is evident that the image of the vertex lies on the line of reflection, so the image of the vertex will be perceived at the intersection of the line of reflection and the normal dropped from the [actual] vertex to the sphere or to the tangent to the circle forming the common section of the sphere’s surface and
the plane of reflection. Furthermore, from any given point facing this [sort of] mirror a cone can be imagined orthogonal to the mirror’s surface or to its extension, and the vertex of this cone is the given point. The line from that point to the image of that point will lie in the plane of reflection as well as on the normal to the mirror’s surface or its extension in the way discussed before, for the point seen and its image always lie in the plane of
reflection, and so does the line extended from the point seen to its image.

2.10◉ In the case of cylindrical mirrors [polished] on the outer surface, the things we claimed about the rod and the cone [in the previous example] are not observed, because in these mirrors what is straight does not appear straight, and there occurs a visual error whose cause we will explain later.⁑ Still, for [any] single point on a visible object [seen in such a mirror] the image-location that has been described can be observed.

2.11◉ For instance, having set up the apparatus [described] in the previous book, insert the panel with the [convex] cylindrical mirror attached in it so that the midline of the portion of the mirror [contained by the panel] lies in the plane of the panel. And do not let this panel pass [below] the bronze plaque, but let it stand orthogonally to it so that the [bottom endpoint of] the line
along the height of the panel stands on the line that bisects the triangle of the bronze plaque. When the panel is set up on this plaque, fill [the space below] with wax and level the wax so that it lies in the same plane as the [top surface of the] plaque, and this is [done] so that the panel can be placed perfectly perpendicular to the plaque.⁑

2.12◉ Then find a pointed ruler, sharpen its edge, and lay this sharpened edge along the midline of the surface of the [cylindrical] ring.⁑ Let it reach along this line [to the mirror], and mark the spot where
it touches the [mirror in the] panel. Then lay a needle along this line, and on its end let a small white object be attached, and do not let the needle reach all the way to the [mirror in the] panel.

2.13◉ Let the eye[s] be placed in the plane of the ruler, and let one of them be closed. The image of the [small white] object [at the end of the needle] will be seen along the line extended from the point marked [on the mirror] to the point of the needle, and that line is of course perpendicular to the surface of the panel, which is tangent to the cylinder [from which the mirror is formed]
along its line of longitude; and that line is perpendicular to the cylinder’s line of longitude, which lies in the plane of the panel and forms the common section of the cylinder’s surface and the plane of reflection. Also the line of [the cylinder’s] longitude and the line perpendicular [to it] lie in the plane of reflection.

2.14◉ Now, if the location of the eye is shifted, and if it moves along [the outer edge of the upper] surface of the [cylindrical] ring, then, just as before, the [small white] object [at the end of the needle], the image of the object, and the [point of the] needle will appear [to lie] on the same line. Moreover, that line is perpendicular to the midline along the length of the cylinder,
and this perpendicular lies in the plane of reflection, for the [top] surface of the [cylindrical] ring intersects the cylinder along a circle that is parallel to the base of the cylinder, and the eye lies in the plane [of this circle]. And later on we shall demonstrate that, when the eye and the visible object lie in a plane parallel to the base of the cylinder, that plane constitutes the plane of reflection.⁑ In this situation, however, the common section of the cylinder’s surface and the plane of reflection is a circle, and the normal upon which the image and the [small white] object are seen falls orthogonally to a line tangent to the circle.⁑

2.15◉ When all this is done, remove the needle and place the sharp-edged ruler along the midline [of the cylindrical ring’s upper surface] so that it reaches the midline along the longitude of the panel, and attach the sharp-edged ruler firmly to the [cylindrical] ring’s [upper] surface with wax. Then remove the panel with the mirror in it and take a pointed ruler and lay its sharpened edge
upon the midline along the length of the panel, and along the sharpened edge draw a line upon the mirror with ink. Then take a moderate-sized wax triangle, one of whose sides is equal to the height of the panel containing the mirror, let it be moderately thick, and let the surface of this triangle be as flat as possible. Then attach the wax triangle firmly to the panel containing the cylinder below the panel’s base, and place the side of the triangle that is equal
to the height of the panel upon the side of the panel’s base. When that is done, the length of this triangle at the base of the [panel containing the] cylinder will be equal to the [height of the] panel, and in order to make the triangle’s surface as flat as the surface of the panel, place the triangle between a panel and a flat surface and compress it until it is quite flat. Place a sharp-edged ruler on the surface of this triangle, and cut off the bottom edge of
this triangle along the edge of the ruler, and this edge will form a straight line. This line will form the base for the panel containing the mirror.

2.16◉ Then place the panel on the surface of the [bronze] plaque in the instrument, and place the edge of its base, which lies along the length formed by the side of the wax triangle, on the line of longitude of the bronze [plaque], as was done before. The plane of the panel containing the mirror will be perpendicular to the bronze plaque, and this plane cuts the bronze plaque along the
line of longitude of the bronze plaque, and this plane is tangent to the surface of the mirror along the line on the mirror’s surface. And this plane is the surface of the panel containing the mirror, and the angle [along the edge] of the pointed ruler will be attached along the midline of the [top] surface of the ring, and the mirror’s surface will be tilted downward to that surface on the side of the triangle’s vertex, since one side of the panel has been raised
according to the breadth of the triangle, while the other side [of the triangle] beyond the vertex of the triangle is the plane of the bronze [plaque], and the line that lies along the midline of the mirror will be inclined.

2.17◉ And when the side of the wax triangle is [positioned] along the line of longitude of the bronze [plaque], the panel that contains the mirror will move along the line of longitude of the bronze plaque, and the [base] side of the [wax] triangle will move with it, if it is [positioned] on the [bronze plaque’s] line of longitude. And let it move back and forth until the point of the
sharp-edged ruler meets some point on the [mid]line on the surface of the mirror and the mirror fits snugly against the sharp-edged ruler, and erase the line drawn in ink on the mirror. Mark the point on the surface of the mirror that is directly under the end of the sharp-edged ruler. And remove the sharp-edged ruler, and apply the needle, and let the needle lie along the midline of the [top] surface of the ring, and let it be attached firmly with wax. And the
imaginary line from the point of the needle to the point marked on the surface of the mirror will be perpendicular to the plane of the panel which is tangent to the mirror’s surface at the marked point, and it will lie perpendicular to any line extended through that point in the plane tangent to the mirror. Thus, it will be perpendicular to the straight line tangent to the common section of the top surface of the ring and the [mirror’s] surface.

2.18◉ Now, place the eye in the plane of the ring at its apex [on the midline along its top surface], and it will look into the mirror until it perceives the form of the small [white] object on the needle, and then it will perceive that object, the point marked on the mirror, and the image of that object [all in a line]. And the line passing through the small [white] object and the point
marked on the [mirror’s] surface is perpendicular to the plane tangent to the mirror’s surface at the point marked on it. And this surface on the ring is among the planes of reflection, and the small [white] object and the centers of sight lie in that plane, and the point of reflection is in that plane, and we will prove this later on. And in this case the image of the small [white] object will lie upon a straight line extended from the small object [itself]
directly to the plane tangent to the mirror’s surface, and since this line is perpendicular to the straight line tangent to the common section of the surface of the mirror and the plane of reflection, which is the [top] surface of the ring, and since the plane of reflection is among the oblique planes that cut the cylinder between the lines of longitude on the cylinder and the circles parallel to its bases, then, because the panel and the mirror in it are at a
slant, the common section of this plane [of reflection] and the surface of the mirror is among the cylindrical sections [i.e., an ellipse].⁑ And we will explain the image-location in this same way if the orientation of the panel containing the mirror is changed and slanted on its surface at a smaller or greater obliquity.

2.19◉ It is therefore evident from these things that the image is perceived where the normal extended from the visible point to the surface of the mirror intersects the line of reflection, and this is the predicted location. If lines are drawn from the visible point to the mirror’s surface, the one that is perpendicular is shorter than any of the others, for any of the others first
intersects the common section of the plane tangent to the mirror’s surface to which the perpendicular falls and the plane of reflection that reaches the mirror, and any line extending in this plane from the visible point to this common section is longer than the perpendicular, since it subtends a greater angle, so what was proposed [is demonstrated].⁑

2.20◉ The same procedure can be applied in the case of the conical mirror [polished] on the outer surface, and the same thing will be shown whether the images of objects are seen in [planes formed by] conic sections, or whether they are seen in [planes formed by] cuts made along lines of longitude.⁑

2.21◉ In concave spherical mirrors some images are perceived behind the mirror, some on the mirror’s surface, and some in front of the mirror, and some of these images are perceived according to reality, whereas some are perceived in ways that do not accord with reality.⁑

2.22◉ That all of those images seen according to reality appear at the point where the normal [dropped from the object] intersects the line of reflection will be demonstrated as follows. Fashion a cone that is orthogonal to its base [i.e., a right cone], and let the diameter of the base be smaller than the radius of the sphere [from which the concave mirror is formed], and let the line
along the length of the cone be longer than that radius. Now, [from the vertex] toward the base, cut off a length [along the cone’s line of longitude] equal to that radius, inscribe a circular section [parallel to the cone’s base at that point] and cut the cone according to this circle. Then, in the middle of the mirror draw a circle the same size as the base of the cone that is left [after the cut], fit the cone to this circle, and attach it firmly with wax.

2.23◉ Place the center of sight where the image of the cone can be perceived, and direct light upon it so that the perception may be clearer.⁑ You will not see a cone conjoined with this one, but you will perceive [the image of]
this [cone] extending behind the mirror, so a certain cone directly in line [with it] will appear with its base behind the mirror along with the wax section of its cone.⁑ And if a line of longitude is marked with ink on this cone, this line will appear to
extend along the surface of the cone that appears [in the mirror], and since the vertex of the cone lies at the center of the sphere, the line extending from the vertex along the length of the cone will be perpendicular to the tangent to any circle on the sphere that passes through the endpoint of the line.

2.24◉ Hence, any line of longitude on the cone that appears [in the mirror] is perpendicular to the line tangent to the common section of the plane of reflection and the surface of the sphere, that line being the common section, and it is a [great] circle. And any point on the cone [lying] on this perpendicular is seen, and every [such] perpendicular lies in the plane of reflection, because
the visible point and its image lie on the normal as well as in this plane of reflection. But every image is perceived along the line of reflection, so the image of any point on the cone will lie at the point of intersection of the normal and the line of reflection.

2.25◉ Furthemore, points whose images are perceived in front of the mirror—i.e., ones that lie between the center of sight and the mirror—[are perceived this way] since the line extended from any of them to the center of the mirror will cut the side of the line lying between the visible object and the mirror. In order to observe this [fact], remove the cone from the center of the mirror and
replace it at the side. The vertex will be the center of the mirror, and let the distance between the center of sight and the mirror be greater than the sphere’s radius. Then take a thin white rod, and stand it on the mirror so that the center of the mirror lies directly between the top of the rod and the center of sight, and direct your line of sight to the point on the mirror from which the line extended to the vertex of the cone lies between the top of the rod
and the center of sight. Then, look into the mirror until neither the top of the rod nor the rod itself appears, and the form of the top of the rod will appear above the mirror and nearer to the eye than the vertex of the cone. And the vertex of the cone, the top of the rod, and the image of the top of the rod will lie on the same straight line, and this line is perpendicular to the line tangent to the common section of the mirror’s surface and the plane of
reflection. For the plane of reflection passes through the center [of sight] and the visible point, and the line passing through these two points lies in the plane of reflection, and the common section is a circle. And this line will form a diameter in this circle, because the center of that circle is the center of the sphere, so this line will be perpendicular to the line tangent to the circle at the endpoint of this line, and this line passes through the visible
point as well as through its image. And so, any point [whose image is seen] in front of the mirror is perceived on the same line as the center [of the sphere] and the image, and any point is seen on the line of reflection, so [it is seen] at the point of intersection of the normal and the line of reflection.

2.26◉ And the things that are perceived in these mirrors as they exist in reality are those whose images appear behind the mirror or in front of the mirror, and beyond these [types of images] there are none that sight may perceive as they exist in reality in this [sort of] mirror, for these latter types of objects do not permit their true images to appear. Images that appear on the surface
of this [type of] mirror belong to the last category, and we will explain this when the discussion turns to visual errors. Therefore, any point that is perceived as it actually exists in this [sort of] mirror appears at the intersection of the normal and the line of reflection, and this normal passes from the visible point to the center of the sphere and falls orthogonally on the tangent to the common section [of the mirror and the plane of reflection].⁑

2.27◉ In the case of concave cylindrical mirrors the image varies, for its location will sometimes be on the surface of the mirror, sometimes behind it, and sometimes in front of it. And in all these cases [the image] is sometimes perceived as it actually exists, sometimes not.

2.28◉ If you want to observe the image-location in these cases, do what you did [before] in the case of cylindrical mirrors [polished on the] outer surface. Insert the panel with the concave cylindrical mirror on it into the ring, as it was inserted earlier, and likewise [position] the needle and the small [white] object at the end of the needle. Place your eye directly in front of the
center of the circle [passing through the middle of the mirror] and at the middle of the [top] surface of the ring, raise the eye a little bit above the ring’s [top] surface, and let it look until it sees the image of the object and perceives the form of the object, the object itself, and the point marked on the mirror along the same line that is perpendicular to the mirror’s surface—and [determine] this according to sense-induction.⁑ The image will lie behind the mirror, and the reflection will occur from a point on the line [of longitude] that lies in the middle of the mirror.

2.29◉ Then place the center of sight in the plane of the [top surface of the] ring, but outside the midline, until it sees the image of the small [white] object [on the needle]. You will see it in front of the mirror, and you will see the object, its image, and the point marked on the mirror along one straight line perpendicular to the straight line tangent to the circle that is parallel to
the base of the mirror and [lying] upon the point marked on the mirror’s surface. And the plane of this circle forms the plane of reflection in this case, and it is the top surface of the ring, and the point of reflection is a point on that circle.⁑

2.30◉ Afterwards, with your other hand place another needle with a small object attached to its end, and lay it on the [top] surface [of the ring] and the axis in such a way that the object and the point marked [on the mirror] lie on the same line according to sense-induction. Let the eye lie in the plane of the [top surface of the] ring between its endpoint and the middle [of the ring’s
top surface]. It will see the image of the object, and it will see this image as well as its [generating] object along with the point marked on the surface of the mirror on the same straight line.⁑

2.31◉ If, however, the straight line [of longitude in the middle of the panel] is slanted according to the small triangle we fashioned—and the center of sight should be on the midline of the ring—you will see the image in front of the mirror, but on the same straight line as the body and the point marked [on the mirror’s surface]. And this reflection will occur from one of the cylindric
sections [i.e., ellipses], because the mirror is inclined, and we know that an image is perceived only along a line of reflection. Therefore, it is clear that the image-location lies where the normal intersects the aforementioned line of reflection, since the [image] is perceived properly, and even though the image may not be perceived clearly, these images will still belong to the category of those that are properly seen.

2.32◉ In the same way you can observe [the image] in a concave conical mirror at the intersection of the normal with the lines of reflection. It is therefore evident that in all mirrors images are perceived at the aforementioned location, which is likewise referred to as the »image.«

2.33◉ We will now explain why visible objects are perceived through reflection where the image is located and why the image lies on the normal [dropped] from the visible object to the surface of the mirror. When the visual faculty grasps a form by reflection, it grasps it immediately without an accurate determination, and it grasps its distance through estimation. It may perceive this
distance accurately through a close and careful scrutiny, or it may not. And we have explained this [process] in the second book, where it was claimed that the visual faculty grasps distance through a deduction based on the size of the object and the particular angle under which the size is perceived. The perception of a visible object whose location is unknown clearly occurs in this way. Also, objects [whose locations are] known are perceived in this way, for
they are compared to known objects and known sizes or distances.⁑ When sight perceives some object through reflection, it perceives the distance of the image solely by estimation; then, after a close scrutiny, it grasps the distance and verifies [it] deductively on the basis of the size of the visible object and
the angle of the cone according to which the form is reflected to the eye.

2.34◉ Thus, if the visible object is among things familiar [to the perceiver], the visual faculty grasps its distance according to an already-known distance [occupied by such a body] subtending an angle equal to this one and [lying at] a distance similar to this one. Likewise, when the visible object is unknown, its magnitude will be compared to another magnitude belonging to familiar
objects, and the distance of this image is grasped according to a deduction based on the size of the angle that the image subtends at the center of sight at the time of reflection. Moreover, the place where the form of the visible object is perceived through reflection [is such that] the form reaching the eye directly from it at a [given visual] angle will arrive according to the very same cone according to which the form is reflected to the eye, and the same cone
will encompass the entire form as it does at the image-location. Hence, when it grasps the visible object through reflection, the visual faculty grasps it where the image lies, for, because the form perceived through reflection at the image-location is identical to the form as perceived directly, it is contained by that [same] cone, and this is why it is perceived at the image-location.⁑

2.35◉ We will now explain why the image will be perceived on the normal. We know that a point that is perceptible to sight exists not intellectually but sensibly, and its form is sensible. I say therefore that in the case of plane mirrors, since the image does not appear on the surface of the mirror but behind it, it is more appropriate and reasonable for it to appear upon rather than
outside the normal. For, since it has been defined according to a spot on the normal, the image’s distance from the point of reflection on the mirror—which is a segment of the line of reflection extended from the location of the image to the point of reflection—is equal to the distance of the point that is seen from the point of reflection. Since the surface of the mirror is orthogonal to the normal, then the line extended [on that surface] from the point of
reflection to the normal forms a side common to the two triangles, and it will bisect the normal. Hence, two sides of one of the triangles will be equal to two sides of the other, and an angle [of one will be equal] to an angle [of the other], for both are right, so the base [of the one will be equal] to the base of the other.⁑

2.36◉ Thus, if the image appears on the normal, it will lie the same distance from the point [of reflection] and the eye as the object from which it originates, and the image will occupy an equivalent location with respect to the point of reflection as the point that is seen does with respect to the same point [of reflection], and the location [of both the image-point and object-point] will
be equivalent with respect to the center of sight, so in this situation both the point seen and its image will appear according to reality. If, however, the image lay beyond the normal, then, since it must lie on the line of reflection, it will lie either beyond or in front of the normal with respect to the eye. If the image lies beyond, then it will lie farther from the point of reflection and from the eye than the point seen, so it will subtend a smaller angle
in the eye than the point [itself]. Moreover, it occupies a smaller area on the eye, so, when it should be equal, it will appear smaller than it[s generating object]. On the other hand, if it lies in front of the normal, it will appear larger [than its generating object-point] because it lies nearer.⁑

2.37◉ In the case of a convex spherical mirror, the image appears on the normal, for the image that appears is either of the center of the eye or of some other point.⁑ If the image is of the center [of the eye], then I say that it is more appropriate for the image of the center of the eye to appear upon the normal dropped from the [center of the] eye to the center of the sphere
than [to appear] anywhere else, for if the form propagates directly along this normal to the center of the sphere, then it will always maintain a uniform disposition with respect to the eye, and so the form that faces any point on the sphere and moves orthogonally toward the center will maintain a uniform disposition with respect to the eye. And the disposition of the form will be the same on one normal as it is on another, for the center of the sphere has the
same situation with respect to every point on the sphere, and all normals of this sort are identically situated.

2.38◉ On the other hand, if the image propagates to some point on the sphere outside the normal, its disposition with respect to the eye will change, for it will have a different situation outside the normal than it does on the normal itself, and the normal moves outside the mirror rather than inside it. But if it appears outside the normal, it will not maintain a [uniform] disposition, yet
it was [agreed earlier that it is] more fitting for the image to maintain [a uniform] disposition than for it to change [its situation] if the visual faculty is to perceive the visible object with certainty. Accordingly, the image of the [eye’s] center appears on the normal, yet we cannot [on that account] determine a specific point for that image on the normal, because there is nothing to be found in any point on the normal to give it precedence over any other
point in determining where the image should appear on it. But we know that at whatever point on this normal the image [of the eye’s center] appears, it invariably appears continuous with the [whole] eye appearing [in the mirror], and it always maintains [the same] location and situation with respect to the entire form [of the eye] that is seen.

2.39◉ The image of any point on the eye other than the center reaches [the mirror] obliquely, so it does not maintain a uniform disposition with respect to the object-point that is seen [i.e., the center of sight], and the normal dropped from the point seen to the mirror[’s surface] falls to the center of the sphere, and it is upon that normal that the image maintains a uniform situation.
There is thus no point where the perceived image will maintain a uniform disposition except upon that normal, and since it must be perceived along the line of reflection, it will be perceived at the intersection of this line with that normal.⁑ We have thus explained the reason for this phenomenon, but the [essential] character of natural objects is related to the [essential] character of their [underlying] principles, and the [underlying] principles of natural objects are hidden.⁑

2.40◉ The same way of proving this will obtain in the case of a concave spherical mirror, as well as in the case of a concave or convex conical [mirror]. In general, for any kind of mirror, image-location will be on the normal, for there is nowhere outside the normal where the form will maintain a uniform and equivalent disposition.

2.42◉ We say that, for each and every point that is perceived by sight in a plane mirror, when it lies outside the normal that falls from the center of sight to the surface of the plane mirror, the line along which the form of that point is reflected to the center of sight will intersect the normal dropped to the mirror’s surface from that point. Moreover, this intersection-point, which
constitutes the image-location [of the object-point], lies behind the mirror[‘s surface], and this [image] will lie the same distance from the mirror’s surface as the [object-]point viewed [in the mirror]. And the visual faculty perceives the image of the point viewed [in the mirror] only in that location, and only one image of any point perceived by sight in such a mirror will be seen.

2.43◉ Moreover, for any point that sight perceives in a convex [spherical] mirror, when its form radiates outside the normal dropped from the center of sight to the center of the mirror, the line along which the image is reflected to the eye will intersect the line extended from that point to the center of the mirror, that line being the normal dropped from that point, and it is orthogonal
to the line tangent to the common section of the plane of reflection and the surface of the mirror. Moreover, where the point of intersection (which constitutes the image-location) lies with respect to the mirror’s surface will depend upon where the center of sight lies with respect to the mirror’s surface. The intersection-point may lie beyond the mirror, on the mirror’s surface, or behind the mirror[‘s surface]. But the visual faculty perceives all [these]
images behind the mirror, even though their locations may vary, and it does not perceive the location of any image on the mirror’s surface through deduction. Also, no matter what point is perceived by sight in this sort of mirror, it yields only one image.⁑

2.44◉ In the case of a convex cylindrical mirror (and the same holds for a convex conical mirror), if any given point that is perceived by sight lies outside the normal dropped from the center of sight orthogonal to the plane tangent to the mirror’s surface, the line along which the form is reflected to the center of sight intersects the normal dropped directly from that point to the line
tangent to the common section of the plane of reflection and the surface of the mirror. And some of the images in these mirrors lie beyond the mirror’s surface, some on the surface itself, and some inside it.⁑ But the visual faculty grasps all of the images in these mirrors behind the mirror’s surface, and whatever point the visual faculty may perceive in these mirrors produces only one image.

2.45◉ In the case of a concave spherical mirror, some of the lines along which the forms of visible points are reflected intersect the normals extended straight from those points to the lines tangent to the common sections of the surface of the mirror and the plane of reflection, and some are parallel to these normals. For those that intersect the normals, the point of intersection, which
constitutes the image-location, sometimes lies behind the mirror outside the point of intersection—and this perception is unclear.⁑ Moreover, of the visible objects that sight grasps in this [sort of mirror], some yield one image only, some two, some three, and some four, but there can be no more [than four].

2.46◉ In a concave conical or a concave cylindrical mirror, some of the lines along which forms are reflected intersect the normals extended straight from the visible points to the lines tangent to the common sections [of the mirror’s surface and the plane of reflection], and some are parallel to the normals. In the case of those that intersect the normals, the intersection for some of the
points lies behind the mirror, and the intersection for others lies in front of the mirror. Some of those that lie in front of the mirror will lie between the mirror and the eye, some at the center of sight [itself], and some beyond the center of sight. The perception by sight of visible objects in this [sort of mirror] sometimes occurs at the [proper] image-location, which is the point of intersection, and sometimes outside the point of intersection. And of those
objects that are perceived [in this sort of mirror], one yields a single image only, another two images, another three, and another four, but there is no way that a thing can yield more than four. And we will demonstrate all these points theorematically.

2.47◉ [PROPOSITION 1] Let A [figure 5.2.1, p. 220] be the object-point, B the center of sight, and DGH a plane mirror. Let G be the point of reflection and DGH the common section of the plane of reflection and the mirror’s surface. From point G draw EG perpendicular to that common section. From point A draw AH perpendicular to the mirror’s surface, and continue it behind the mirror. Let AG
be the line along which the form [of point A] reaches the mirror and BG the line along which it is reflected to the center of sight. Accordingly, BG, EG, and AG lie in the plane of reflection, and since [normal] EG is parallel to [normal] AH, while BG is oblique to EG, BG will intersect AH. Let it then intersect at point Z. I say that ZH = HA.

2.49◉ Now, if we wish to determine where the point of reflection lies according to normal [AHZ], let a segment along the normal be cut off below the mirror equal to the segment on it [from object-point A] to the mirror, i.e., so that ZH = AH. Then extend line BGZ from the center of sight to point Z. I say that G is the point of reflection.

2.51◉ [PROPOSITION 2] Furthermore, let A [figure 5.2.2, p. 220] be the center of sight, let AG be the normal to the plane mirror, and let D [be the point] on the eye’s surface [where it] intersects this normal. I say that there is no point other than D on this normal that may reflect from this mirror to the eye.

2.52◉ For if [some such] point may be taken on this normal beyond the eye, let it be H. Its form will not reach the mirror along the normal because of the interference of the opaque body [of the eyeball], and so its form is not reflected along the normal.

2.53◉ But if it is claimed that [its form] can be reflected from some other point on the mirror, let that [point] be B. Its form will propagate to point B along line HB, and it is reflected along line BA. Let angle HBA be bisected by line TB [so that HBT is the angle of incidence and TBA the angle of reflection]. Therefore, TB will be normal to the mirror’s surface. But TG is normal to that
same [surface], so two perpendiculars will have been dropped from the same point [T] to the mirror’s surface, which is impossible.

2.54◉ The demonstration that the form of point D cannot be reflected from any point other than G on the mirror will be the same, so that form is reflected only along the normal. If, however, it is claimed that the form of any point selected between G and D on this normal is conveyed to the center of sight by reflection, the disproof follows from the fact that the body [of the selected
point] will be either opaque or transparent.

2.55◉ If it is opaque, then the point’s form will proceed to the mirror along the normal and will return along the same line to the center of sight, but because of the [point’s] opacity that form cannot pass through to reach the center of sight.

2.56◉ On the other hand, if that point is transparent, its form will mingle with it and fuse with it upon its return from the mirror along the normal, and it will not be reflected to the center of sight.

2.57◉ Moreover, by the preceding method, it can be proven that the form of any point selected on this normal between G and D cannot be reflected to the center of sight from any point on the mirror other [than G]. By the same token, the form of [any] point selected between A and D is reflected to the center of sight neither along the normal nor along any other line, for the points lying
between the center of the eye and its surface are absolutely transparent, so their form is neither conveyed back nor reflected in such a way as to be sensed. And since no point other than D selected on the surface of the eye faces the mirror at a right angle, any such point will be seen upon the normal dropped from it to the mirror, and its image [will lie] the same distance behind the mirror’s surface as the point itself [lies above it]. And since D appears
continuous with the other points on the surface of the eye, and since its image appears continuous with the other images [of those points], the image of D will appear [to lie] the same distance from the mirror’s surface as D lies from it.⁑

2.58◉ Hence, it is evident that the image of any point seen in the mirror will appear on the normal, and the distance of the image from the mirror’s surface and [the distance] of the visible body [from the mirror’s surface] is the same.

2.59◉ [PROPOSITION 3] Furthermore, the form of a point viewed in a plane mirror is reflected to the same center of sight from only one point. Let A [figure 5.2.3, p. 220] be the center of sight, B the point viewed, and ZH the mirror. If, then, it is claimed that the form of B is reflected to the center of sight from two points on the mirror, let one of them be point D, the other E. Draw
line AB from the point viewed to the center of sight; this line will either be perpendicular to the mirror or not.

2.60◉ If it is not perpendicular, we know that the line lies in the plane of reflection [which is] orthogonal to the surface of the mirror, and it will lie in only one such plane. For if [it lies] in two, it will be common to two orthogonal planes, and if a point is chosen on it and a line is extended from it in either plane to the common section of that plane and the surface of the mirror,
this line will certainly be orthogonal to the mirror. Likewise, from that [same] point extend a line in the other plane to the common section of that plane and the surface of the mirror, and this line will be orthogonal to the mirror, so from the same point two perpendiculars will have been dropped [to the same plane].⁑

2.61◉ Thus, since BA lies in only one orthogonal plane, and since the three points A, B, and E lie in the same orthogonal plane, AE and EB will also lie in the orthogonal plane that contains AB, and so will EB and DB. Therefore, EA and EB [will lie] in the same plane as DA and DB. But angle AEH = angle DEB [by supposition], and angle HEA > angle ADE, because it is an exterior [angle of
triangle AED], so [angle] BED > [angle] ADE. But [angle] BDZ = [angle] ADE [by supposition], while [exterior angle] BDZ [of triangle BED] > [interior angle] BED, so [angle] ADE [which = angle BDZ, by supposition] > [angle] BED, but it was [just] claimed that it is smaller. It follows, therefore, that reflection may occur from only one point.

2.62◉ On the other hand, if AB is perpendicular to the mirror, then it has already been claimed [in proposition 2] that there is only one point on the line dropped orthogonally from the center of sight to the mirror whose form is reflected from the mirror to the center of sight. And it has already been demonstrated that the image of that point reflects from only one point [on the mirror],
so what was proposed [has been demonstrated].

2.63◉ [PROPOSITION 4] Furthermore, when any point is viewed by both eyes, one and the same image appears to both, and it does so in the aforementioned [image]-location. Now, it is obvious that the form of the point does not reflect to both eyes from the same point on the mirror. For if the line of reflection proceeding to one eye were to form the same angle with the normal erected to the
mirror’s surface [at the point of reflection] as the line along which the form reaches the mirror, then another line could not be chosen in the same plane to form with the normal an angle equal to this one.⁑ Hence, from this point no [form
incident along the same] line will be reflected to the other eye. Reflection must therefore occur from different points on the mirror.

2.64◉ Let those points be T and Z [figure 5.2.4, p. 221]. Let QE be the plane mirror, A the point viewed, B and G the two centers of sight, and AD the normal [dropped from object-point A to the mirror]. It is evident, then, that BT, AT, [segment DT of] ET, and AD lie in the same plane orthogonal to the surface of the mirror. Likewise, AD, AZ, and GZ lie in the same orthogonal plane, and DT
is the common section of plane ADTB [and the mirror’s surface], while DZ is the common section of plane ADZG [and the mirror’s surface]. If BT and GZ lie in the same orthogonal plane, TDZ will form a single line, and normal AD will lie between the two aforementioned perpendiculars [dropped] to the mirror’s surface from the two centers of sight [i.e., the perpendiculars dropped from B and G to the plane of the mirror containing line QDE], or it will lie outside
them [figure 5.2.4a, p. 221].

2.65◉ Whichever the case, line BT of reflection will cut from normal AD a segment behind the mirror equal to segment AD [above it, as demonstrated in proposition 1 above]. Likewise, GZ will cut from the same normal a segment behind the mirror equal to that segment [AD above the mirror]. Those two lines of reflection will therefore intersect the normal at the same point [H] behind the
mirror. Hence, the image of point A will be perceived by both eyes at the same point on the normal, so there will only be one image, and it will be the same [for both eyes], and it will lie at the same place as it would if it were viewed by only one eye.

2.66◉ If, however, points T and Z do not lie in the same plane orthogonal to the mirror [figure 5.2.4b, p. 221], precisely the same proof will apply, since each line of reflection cuts a segment on the normal [below the mirror] equal to the segment above it, and the intersection of the lines of reflection with the normal will be at the same point, so what has been proposed [is
demonstrated].

2.67◉ On the other hand, if point A lies on the normal dropped to the surface of the mirror from only one center of sight, it is perceived by that same center of sight behind the mirror at a point on the normal that lies as far from the mirror’s surface [below it] as A lies from it, so the form of A appears continuous with the forms of other points that appear in proximate locations. Also,
the image of A is perceived by the other eye at the same point on the normal, and so only one image of point A is seen by both eyes, and [it appears] at the same point on the same normal, which is what was proposed.⁑

2.68◉ [PROPOSITION 5] What we claim will be clear in the case of spherical convex mirrors. Let A [figure 5.2.5, p. 222] be the point seen, B the center of sight, and G the point of reflection. It is clear that BG and AG lie in a plane [of reflection] orthogonal to the plane tangent to the sphere at point G. Let ZGQ be the [great] circle that forms the common section of the plane of
reflection and the surface of the sphere [from which the mirror is formed]. Let PGE be the line tangent to this circle at the point of reflection. Let HG be the normal to this line [at the point of reflection]. It is evident that HG should reach the center of the circle. But if not, then, since the line extended from the center of the sphere to point G is also perpendicular to line PGE, two lines perpendicular to one [and the same] line will have been drawn from
the same point on the same side [of that line].⁑

2.69◉ Now, let N be the center of the sphere, and let a line, i.e., AN, be extended from the point viewed to the center of the sphere, that line being normal to the plane tangent to the sphere at the point on the sphere through which it passes. And since it is manifest that [line of reflection] BG intersects the sphere, because it lies between HG and GP, which form a right angle, it will
intersect line AN. And since normal HG lies in the plane of reflection, the center of the sphere will lie in the same plane, and so [normal] AN [will lie] in the same plane as [normal] HG.

2.70◉ Accordingly, let D be the intersection of [line of reflection] BG with [normal] AN. It is clear that D will be the image-location, and this analysis must be understood [to obtain] when the line extended from the point seen to the center of sight is not perpendicular to the mirror.

2.71◉ [PROPOSITION 6] Now, let line PGE [figure 5.2.6, p. 222] intersect line AN. Let the point of intersection be E, and this point is designated as the endpoint of tangency. I say that in this case the line extending from the center of the sphere to the image-location is longer than the line extending from the image-location to the point of reflection: that is, DN > DG.

2.73◉ Let a line [AH] be drawn from point A parallel to DG, and let it intersect line HN at point H. Accordingly, angle NGD = [alternate] angle GHA. But angle NGD = angle AGH [by previous conclusions]. Therefore, angle GHA = that same angle [AGH], so the two sides AG and HA [of triangle AGH] are equal. Therefore, AH:DG = AG:DG. But AH:DG = AN:DN [by Euclid, VI.4, because triangles AHN and
DGN are similar], so AN:DN = AG:DG. Therefore, AN:AG = DN:DG [by Euclid, V.16]. But AN > AG, because it subtends more than a right angle in triangle ANG [by Euclid, I.19 and 32]. Accordingly, DN > DG, which is what was proposed.

2.74◉ [PROPOSITION 7] I say, further, that the line dropped orthogonally to the sphere from endpoint of tangency E, i.e., the segment [EF] of line EN [between E and the surface of the sphere], is shorter than the radius [of the sphere].

2.77◉ [PROPOSITION 8] Now, let G [figure 5.2.8, p. 223] be the center of sight, D the center of the sphere, and DZG the normal [dropped] from the center of sight to the sphere. I say that the form of no point other than the point on the surface of the eye [through which normal GD passes] is reflected along this normal.

2.78◉ For the forms of the points selected beyond the center of the eye are not reflected for the reason given above [in proposition 2]. And the same holds for the points lying between the surface of the eye and the mirror. I say, as well, that no point on this normal is reflected from any other point on the mirror.

2.79◉ For if it is claimed that [reflection does occur] from another point, let that point be A. Line GA will be the line of reflection, and from that point [on the normal, which is assumed to reflect its form to G, i.e., X] we will imagine a line to A, that line [AX] being the one along which the form proceeds [to the mirror]. These two lines form an angle at A, and diameter DA will
necessarily divide that angle, since it is normal to point A, and the normal bisects the angle produced by the form’s line of incidence and its line of reflection. And so diameter DA will intersect normal GD between the selected point and G [i.e., at point E]. And so the two straight lines [GD and DE, which consists of DA and its rectilinear continuation AE] will intersect at two points [D and E] and will form a plane.

2.80◉ It therefore follows [from the impossibility of such a double intersection] that the form of only the one point that lies on the surface of the eye may be reflected from the mirror along the normal, and it must appear at its original image-location according to its continuity with other [surrounding] points.⁑

2.81◉ [PROPOSITION 9] Furthermore, let GA and GB [figure 5.2.9, p. 224] be lines drawn tangent to the sphere from the center of sight, and mark off the [great] circle upon which the plane formed by these lines cuts the sphere. AB will be the visible portion of this circle. I say, therefore, that some of the image-locations perceived according to reflection from this portion lie inside the
mirror, some on the mirror’s surface, and some outside the mirror. Each one of these cases must be accounted for.

2.82◉ From point G [figure 5.2.9] draw a line that cuts the circle, and let the segment of it that forms the chord on the [intersected] arc of the circle be equal to the circle’s radius. Let that [cutting] line be GHK, and let HK be the [segment on it forming the] chord that is equal to the radius. Then, from point H draw normal DHM. I say that the [image-]location of a form reflected from
point H will lie inside the sphere.

2.83◉ From point H draw a line OH that forms with MH an angle [OHM] equal to angle MHG. Points [incident to the mirror] on this line, and on no other, will be reflected from point H to the center of sight [G]. Accordingly, choose some point O on it, and draw line OD [i.e., the normal dropped from the object-point] from that point to the center of the sphere. OD will be perpendicular to the
plane tangent to the sphere at the point where OD passes through it. But, by construction, angle OHM = angle MHG, so, by the same token, it is equal to [angle MHG’s] alternate angle KHD. But [angle] KHD = [angle] KDH, because they are subtended by equal sides [of equilateral triangle KDH, i.e., radii DH and DK].

2.84◉ Accordingly, angle OHM = angle KDM, so lines KD and OH are parallel. Therefore, if they are extended indefinitely, they will never intersect.⁑ But line OD will intersect the
line [HK] joining KD and OH, and so, no matter what point is chosen on line OH, the line extended from that point to point D [i.e., the normal] will intersect the line of reflection inside the sphere, and that line will be perpendicular to the sphere, as represented by OD [so it will be the normal on which the image must lie]. Hence, the image of any point on line OH will appear inside the sphere [where the line of reflection GH intersects normal OD].

2.85◉ Now, the arc on the circle lying between point H and the point where the normal dropped from the center of sight passes through [the circle] is HZ. I say that, from whatever point on this arc reflection occurs, the image-location will lie inside the sphere.

2.86◉ [Here is] the proof. Let I [figure 5.2.9a, p. 224] be the point selected, draw line GIS from the center of sight to intersect the circle at that point, and draw normal DIT from that point. Make line PI form an angle [PIT] with [normal] IT that is equal to angle TIG. It is clear that only points on line PI are reflected from point I to the center of sight. It is also clear that line IS
> line KH [by Euclid, III.8], so it is longer than [line] SD [which = line KH, by construction]. Therefore, angle SDI > angle SID [by Euclid, I.19] , so it is greater than angle GIT [which is vertical to angle SID], and therefore greater than angle TIP.

2.87◉ Hence, line PI and [line] SD will never intersect [in the direction of P and S],⁑ and the line
extended to point D from any point on line PI will intersect line SI inside the sphere, SI being the line of reflection [continued inside the circle]. And any line drawn [to D] from any point on line PI will be perpendicular to the sphere, just as PD is. Moreover, since the image-location lies at the intersection of the normal [dropped] from the point viewed and the line of reflection, the image of any point on line PI will lie inside the sphere. It is therefore
evident that the appropriate location of all images [produced by reflection from points] within arc HZ will be inside the sphere, which is what was proposed.

2.89◉ Choose some point on this arc, let it be N [figure 5.2.9b, p. 224], and from point G draw line GNQ to intersect the circle. Then draw normal DNF, and draw line EN to form with normal [DNF] an angle [ENF] equal to angle FNG. Since line NQ < line KH [by Euclid, III.8], it is also shorter than line QD [since QD = KH, by construction], and so angle QDN < angle DNQ [by Euclid, I.19],
and therefore smaller than angle GNF [which is vertical to angle DNQ], and therefore also smaller than angle ENF [which = angle GNF, by construction]. Hence, line EN and [line] DQ will intersect. So let the intersection be at point E. It is obvious that line EQD is normal to the sphere, and it intersects line of reflection GNQ at point Q, which is a point on the sphere. Therefore, if reflection occurs at point N, the image of point E will appear at point Q, and
this point lies on the surface of the sphere.

2.90◉ If, however, a point, such as R, is chosen on line EN beyond E, the normal dropped from that point to the center of the sphere, e.g., RD, will intersect line of reflection GNQ beyond point Q [e.g., at point L]. This lies outside the sphere, so the image of any point chosen on line EN beyond E will lie outside the surface of the mirror.

2.91◉ But if some point [such as C] is chosen on line EN in front of point E, the normal dropped from it to the mirror [along CD] will intersect line GNQ inside the sphere, because [it intersects it] at a point that lies between N and Q. Accordingly, the image of any point chosen on line EN between E and N will appear inside the sphere.

2.92◉ The very same proof will hold for any other point chosen on arc BH. And so, for any point on arc BH, one image alone lies on the surface of the mirror, others lie inside the mirror, and others lie outside it. Moreover, what has been demonstrated for arc ZB can be shown in exactly the same way for arc ZA, and the very same demonstration will apply for any [great] circle on the sphere
and the portion [of it] that is selected to face the center of sight and be bisected by normal GD.

2.93◉ Accordingly, if the center of sight remains immobile with normal GD fixed in place, and if line GHK is rotated uniformly about normal [GD as axis], it will cut off a circular portion of the sphere with its rotation, and the image of any point [reflected from any point] within this portion will appear inside the sphere.⁑

2.94◉ On the other hand, if line GB tangent [to the sphere] is rotated uniformly about normal [GD as axis], it will cut off a portion of the sphere larger than the previous one, and from any point on the portion that constitutes the excess of the latter over the former, [one] image reflected [to the eye] will find its location on the surface of the sphere, and of the remaining images, some
[will lie] inside the sphere, and some outside it.⁑

2.95◉ From these facts, we know that in this [sort of] mirror any image appears on a diameter of the sphere [that forms the normal, and that image may appear] inside the sphere, outside it, or on its surface. Moreover, any diameter upon which an image may appear, whether on the surface of the sphere or outside that surface, lies below the point on the sphere where the tangent extended from
the center of sight to the sphere touches the sphere, that point being at the very edge of the visible portion [of the mirror]. We [also] know [from this analysis] that every line of reflection cuts the sphere in two points, the point of reflection and some other one.⁑

2.96◉ [PROPOSITION 10] It remains for us to determine image-locations more specifically. I say that, if a diameter is chosen [on a convex spherical mirror], and if a line is extended to it from the center of sight to cut the sphere so that the segment lying between the point of intersection on the sphere and the point on the diameter to which it reaches is equal to the segment of the
diameter lying between that latter point and the center [of the sphere], that point does not constitute the location of any image.

2.97◉ For instance, let AG [figure 5.2.10, p. 225] be a [great] circle on the sphere, H the center of sight, and ED a diameter, or normal, in the sphere. Let HZ be a line intersecting the sphere at point F and meeting ED at point Z, and let ZF = ZD. I say that Z does not constitute the location of any image [on normal DE].

2.98◉ For it is evident that there is no image-location at any point other than on ED, because the image of any point lies on the diameter extended from it to the center of the sphere. That the image-location of any point on ED does not lie at Z will be demonstrated as follows.

2.100◉ Hence, if they are extended indefinitely, they will never meet. Accordingly, the form of no point on [normal] ED will pass to point F along [line of incidence] QF, and there can be no image-location for any point at point Z unless its form passes to F along line QF.⁑ The same disproof will hold for any diameter that is chosen, so what was proposed [is demonstrated].

2.103◉ [PROPOSITION 11] Furthermore, if a diameter, other than the one imagined [to extend] from the center of sight to the center of the sphere, is taken between the lines drawn tangent to the sphere from the center of sight, and if the point on it we [just] described as the limit of image-locations is determined, I say that the image-locations of points on that diameter lie only at
points on it between the surface of the sphere and the aforementioned limit.

2.104◉ For example, let BZ and BE [figure 5.2.11, p. 225] be the tangents, B the center of sight, A the center of the sphere, BHA the visual axis [which is normal to the mirror from the center of sight], and DA the selected diameter with G [the point at which DA intersects the sphere and T] the limit [of image-locations].⁑ I say that images of points [whose forms are reflected to B] occur only at points between G and T on DA.

2.105◉ That they will not occur at point G or [any point] outside the sphere’s surface is clear from what was said above [in proposition 9], i.e., that the diameter on which the image-location will occur at the surface of the mirror or outside it lies below the point of tangency [E]; and since diameter DA lies between the lines of tangency, there will be no image-location on it either at
or outside the [mirror’s] surface. On the other hand, it will be demonstrated that the image may fall at any point taken between G and T.

2.108◉ Accordingly, let AC and AG [figure 5.2.12, p. 225] be the tangents [defining] the visible portion [of the sphere], A the center of sight, B the center of the sphere, ADBZ the visual axis [which is normal to the mirror from the center of sight], and ZCG the [great] circle on the sphere lying in the plane of the tangent lines. From the center [of the sphere] draw diameter BG to point
[G] of tangency. It is evident that angle ZBG > a right angle, for, since angle BGA in triangle BAG is right, angle GBA < a right angle, so [adjacent angle] ZBG will be greater [than a right angle]. Accordingly, let HBG be a right angle. HB will thus be parallel to line of tangency AG. Hence, when extended [indefinitely] they will never intersect, whereas any diameter between H and G will intersect line AG.

2.109◉ Draw line AMO from point A to intersect the sphere so that chord MO = radius OB, and let diameter BO intersect line AG at point T. I say that there is an image-location at any point on TO, that there is an image-location at no other point on diameter TB, and that O and T are the limit-points for image-locations [on that diameter].⁑

2.110◉ Take [such a] point, let it be K, and draw [line of reflection] ANK to cut the sphere at point N. Then draw normal BNY, and form angle [of incidence FNY] with FN that is equal to angle [of reflection] YNA. It is clear that [line of incidence] FN will not lie between B and T, because [in that case] it must cut either the sphere or tangent AP in two points.⁑ Therefore, the form of point F will propagate along FN to point N and will be reflected to A along line AN, and its image will appear at point K. And the same proof holds for any other point chosen [as an image-location between T and O].

2.111◉ [PROPOSITION 13] I say, further, that no matter what diameter may be chosen within arc OG, it will contain image-locations inside the mirror, as well as one [image] on the surface of the mirror, and other [images] outside the mirror.

2.112◉ Accordingly, take point L [in figure 5.2.13, p. 226], and draw diameter BL until it intersects [line of tangency] AP at point E. Then draw line AL to intersect the sphere at point R. It is obvious [by Euclid, III.8] that RL < LB, since it is shorter than MO, which is equal to the [sphere’s] radius [by construction from the previous proposition]. Accordingly, if from A a line is
drawn to diameter LB such that the segment [of it] lying between the circle and the diameter is equal to the segment of the diameter [extending] from the point where that line meets the diameter to the center [of the circle], it will fall between L and B. For if it were to fall between L and E, then RL > LB, and every line falling between the center [of the sphere] and that equal segment [on LB] would be longer than the segment on the diameter where it
terminates, according to the proof [in proposition 10] devoted to explaining the limit of image[-locations].⁑

2.113◉ So let I be the point where the line[-segment between arc DG and normal BL] equal [to the segment on BL] will fall. I say that there is an image-location at any point on EI [other than E or I]. And the same demonstration will apply [in this case] as applied for TO [in the previous theorem].

2.114◉ Accordingly, the image-locations on diameter EB are divided among those that lie inside the mirror [i.e., between L and I], those that lie outside the mirror [i.e., between E and L], and the single one, i.e., at point L, that lies on the mirror’s surface. And in this way you can demonstrate [the same thing] for any diameter that passes through arc OG.

2.116◉ Take diameter BQ [figure 5.2.14, p. 227], and let it intersect the tangent [AG] at point P. Then draw line ANQ to cut the sphere at point N. It has already been established [by construction in propositions 12 and 13] that MO = [radius] OB, but NQ > MO, so NQ > [radius] QB. Moreover, the line drawn at the circumference [of the sphere] to diameter PB and equal to the segment of
BP lying between it and the center [of the sphere] will not fall between Q and B. For if it were to fall [there], then, according to the previous demonstration, NQ < QB [which contradicts the conclusion drawn above].

2.117◉ It follows, therefore, that the line equal [to the previously defined line] should fall between P and Q. That it may not fall at point P is clear from the fact that angle PGB is a right angle [and so tangent AGP cannot be a line of reflection]. Therefore, PB > PG [because it subtends a larger angle in triangle BGP]. So it will fall in front of P [i.e., between P and the
mirror].

2.118◉ Let the point at which it falls be S [so that the line-segment on AS between S and the point at which AS cuts the circle is equal to SB]. S will therefore constitute the limit for image-locations, and any point between P and S will constitute an image-location, and the proof for this is the same as above.

2.119◉ From these theorems it is evident that all the images [produced] on diameters within arc HO lie outside [the mirror]; on diameter FB, one lies on the surface, at O; and all the rest lie beyond, i.e., in TO; however, of all the images on [any] diameter within arc OG, some lie inside [the mirror], some outside it, and one on its surface.

2.120◉ [PROPOSITION 15] Furthermore, no diameter can be selected in arc HZ [figure 5.2.15, p. 227] that contains an image-location, for no diameter selected there intersects tangent AP [since HB is parallel to AP].

2.121◉ Now, from some point on such a diameter [outside the sphere], draw a line [tangent] to the sphere. It will of course touch [the sphere] within segment GZC, not within segment GDC, unless it cuts the sphere. Accordingly, no form of any point on such a diameter will reach the portion [of the mirror] visible to the eye.

2.122◉ Moreover, what has been claimed for arc GH can be demonstrated in the same way for the portion of arc CZ that corresponds to it. So if an arc equal to arc HZ is taken on the other side of Z, there will be no image-location on any diameter within that arc.

2.123◉ The same method is used in proving this for any circle [on the sphere], so if line HB is rotated [about axis BZ] while angle HBZ is kept constant [throughout its rotation], it will describe with its motion a portion of the sphere within which there is no diameter with an image-location. If, however, HB is held in place [so that angle HBZ remains constant], and if OB is rotated
[about axis BZ], it will describe a portion [of the sphere] within which all of the images [on any diameter lie] outside [the surface, by proposition 14], but on diameter TB one [image lies] on the surface, the rest outside [by proposition 12]. Finally, if arc OG is rotated, it will form a portion [of the sphere] within which some of the images lie on the surface, some outside the sphere, and some inside [by proposition 13].

2.124◉ Yet the visual faculty does not perceive which images lie on the surface of the sphere or which [lie] outside, nor does it determine anything in the process of perceiving them except that they lie behind the visible portion [of the sphere].⁑ At this point, then, the image-locations in these [sorts of] mirrors
have been determined.

2.126◉ Let the [visible] point be B [figure 5.2.16, p. 228], let A be the center of sight, and let A not stand on the normal dropped to the center [of the sphere through the point of reflection]. I say that [the form of] B is reflected to A from only one point on the mirror, and it yields only one image to the eye in this [kind of] mirror.

2.127◉ It is clear that its form can be reflected from some point. Let that point be G, and draw [line of incidence] BG and [line of reflection] AG. Let N be the center of the sphere, draw diameter BN [i.e., the normal dropped from object-point B] to intersect the sphere’s surface at point L, and let the limits of the [visible] portion [of the mirror] facing the center of sight be D and E
[that portion being subtended by tangents AE and AD]. Let line AG intersect normal [BN] at point Q, which is the image-location.

2.128◉ It is obvious that A, N, and B lie in the same plane orthogonal to the sphere. And since all planes that are orthogonal to the sphere and that contain BN intersect along BN, and since only one plane containing line BN can be extended through point A, it is clear that A and B lie in only one plane that is orthogonal to the sphere, not in several [such planes]. Moreover, since the
visible point [B] and [center of sight] A must lie in the same plane that is orthogonal [to the sphere] at the point of reflection, it is clear that the reflection of [the form of] point B to the eye will occur only on the [great] circle that lies in plane ANB within the sphere. Accordingly, let that circle be DGE. I say, again, that reflection will occur from no point other than G on this circle.

2.130◉ Likewise, reflection cannot occur from any point on [arc] GL. Take some [such] point, let it be Z, and draw line AZO to intersect normal [BN] at point O. Then draw the line tangent to the circle at point Z, a line that necessarily falls between [i.e., intersects] BG and BL, and let it be MZ. Let FG be the line tangent to the circle at point G. It is clear from the above that BN:NQ =
BF:FQ [from proposition 7 above, F being the endpoint of tangency]. So too, BN:NO = BM:MO [from proposition 7 above, Z being the point of reflection and M the endpoint of tangency]. But BN:NQ > BN:NO [by Euclid, V.8]. Therefore BF:FQ > BM:MO, which is clearly impossible, since BF < BM, while FQ > MO. It therefore follows that reflection may not occur from point Z.

2.132◉ Moreover, if point B lies on the visual axis [which is the normal dropped from the eye to centerpoint N of the sphere], it is obvious that it is reflected from only one point, because [it is reflected] only along the normal, and its image will be unique and will lie in the appropriate image-location because of its continuity with the other points [as established in proposition 8].

2.133◉ [PROPOSITION 17] Furthermore, if two points are taken on a given diameter on the same side of the [sphere’s] center, the image-location of the point nearer the [sphere’s] center will lie farther from the center of the sphere than the image-location of the point farther from the sphere’s center. Also, the point of reflection for the point nearer the [sphere’s] center will lie farther
from the center of sight than the [point of reflection] for the point farther from the center of the sphere.

2.134◉ I say, for example, that the image-location for point C [figure 5.2.17, p. 229] lies farther from the center [N] than does the image-location for point B, and that the point of reflection for point C lies farther from point A [the center of sight] than the point of reflection for point B, i.e., point G. I say [further] that [the form of] point C is reflected [to A] from a point on
arc GL only.

2.135◉ For it is evident [from the previous theorem] that it will not be reflected from any point on arc LE, nor from point L, nor from point G, since [the form of] B is reflected from that point. But if it is claimed that [it is reflected] from some point on arc GD, let that point be T, and let CT be the line along which the form [of C] radiates to the mirror. Then draw normal NT, which
will necessarily bisect angle CTA, and draw normal NGK. Angle NTA > [angle] NGA [by Euclid, I.21]. It therefore follows that angle PTA [adjacent to angle NTA] < angle KGA [adjacent to angle NGA], so angle CTP < angle BGK [because, by supposition, angle CTP = angle PTA, and angle BGK = angle KGA, and we have just concluded that angle PTA < angle KGA]. But [by Euclid, I.32] angle CTP = angle TNC + angle TCN, because it is an exterior [angle of triangle
TCN]. In addition, angle BGK [which is an exterior angle of triangle BGN] = angle GNB + angle GBN. Accordingly, the [sum of the] two angles TNC and TCN < the [sum of the] two angles GBN and GNB , which is impossible, since angle TNC contains [angle] GNB as a part, while angle TCN > [angle] GBN.

2.136◉ Hence, it follows that point C may only be reflected from [some point among the] points lying between G and L. All lines drawn from point A to normal BN through such points fall on points [between L and Q] that lie farther than point Q from the center of the sphere, and they fall to points on the sphere [between L and G] that are farther away from the center of sight [A] than point
G, and so what was proposed [has been proven].

2.138◉ Let B [figure 5.2.18, p. 229] be the visible point, [let] A [be] the center of sight, and draw two lines from these points to the center of the mirror. If those lines are equal, it will be easy to find [the point of reflection]. For a [great] circle on the sphere will be defined in the plane within which those two lines lie, and we know that reflection occurs from only one point on
that circle. Accordingly, the angle those two lines form at the [sphere’s] center will be bisected.

2.139◉ Draw the line [GN] that bisects the angle [and extend it] beyond the sphere. It will of course be perpendicular to the line tangent to this circle at the point [G] through which it passes. And if two lines [AG and BG] are drawn to that point, one from the center of sight and the other from the visible point, they will form two triangles with that normal and the two original lines,
two sides of these triangles [i.e., BN, NG and AN, NG] being equal [respectively] and angle [BNG of the one, being equal] to angle [ANG of the other].⁑ Hence, the point on the circle through
which that normal passes constitutes the point of reflection, which is what was set out [to be demonstrated].

2.140◉ On the other hand, if the line extending from the visible point to the center of the sphere is not the same length as the line extending from the center of sight to that same centerpoint, we must set forth some preliminary things, one of which is as follows.

2.141◉ [PROPOSITION 19, LEMMA 1] If a diameter is taken in a circle, and if a point is taken on its circumference, a line can be drawn from that point to the extension of the diameter beyond [the circle] such that its extension from the point where it intersects the circle to the point where it intersects the diameter is equal to a given line.

2.142◉ For instance, let QE [figure 5.2.19, p. 230] be the given line, GB the diameter of circle ABG, and A the given point [on its circumference]. I say that I can draw a line from point A such that [the segment extending] from the point where it intersects the circle to diameter GB is equal to line QE, which will be proven as follows. Draw the two lines AB and AG, which are either equal
or unequal.

2.143◉ Let them be equal, and add a line to QE—let EZ be this additional line—such that the rectangle that will be formed by the whole line augmented by the additional line and the additional line [i.e., (QE + EZ),EZ = QZ,EZ] = AG2.⁑ Accordingly, since QZ,EZ = AG2, QZ > AG. For if EZ were equal to, or longer than AG, it is impossible that QZ,EZ = AG2 [because the whole, QZ, is greater than its part, EZ, whose square = AG2]. On the other hand, if it is shorter, it is clear that QZ > AG.

2.144◉ Accordingly, extend AG [figure 5.2.19a, p. 230] until it is equal [to QZ], and let the result be AGT. Then, placing a compass-point at A, draw a circle according to radius AGT, and [since] this circle will intersect diameter BG, let it intersect [it] at point D. Then draw line AD, which will necessarily intersect circle [AGB], for if it were tangent at point A, it would be parallel
to BG and would never intersect it. Accordingly, let it intersect [the circle] at point H, and draw line GH.

2.145◉ It is clear that, since ABGH constitutes a quadrilateral within the circle, the two opposite angles, i.e., ABG and AHG, sum up to two right angles [by Euclid, III.22]. But [angle] AGB = [angle] ABG, since they are subtended by equal sides, according to construction. Therefore, angle AHG = angle DGA [because, as previously concluded, angle AHG = two right angles – angle ABG, and
angle DGA = two right angles – angle AGB, which = angle ABG], and angle HAG is common to the whole triangle ADG and triangle AHG that forms part of it. It therefore follows that angle HDG = angle HGA and that [the one] triangle [DGA] is similar to [the other] triangle [AHG], so [by Euclid, VI.4] DA:AG = AG:AH. Hence, DA,HA = AG2 [by Euclid, VI.17]. But DA = TA [since they are both radii of circle TD], so DA = QZ [since TA = QZ, by construction]. And [therefore] AH
= EZ, and DH = QE, which is the given line, and so what was proposed [has been demonstrated].⁑

2.146◉ Now, if AB and AG are not equal [figure 5.2.19c, p. 231], draw a line from point G that is parallel to AB, let it be GN, and take some line ZT, and form an angle at point Z with line ZF [i.e., angle TZF] equal to angle AGD. Then, from point T draw a line parallel to ZF, let it be TM, and from angle TZF cut off an angle with line ZM [i.e., angle MZT] equal to angle NGD, for this line
[MZ] will necessarily intersect TM, since it lies between parallels [TM and ZF]. Let the point of intersection be M. It follows, therefore, that angle MZF = angle AGN [since, by construction, angle TZF = angle AGD, and angle TZM = angle NGD, so angle MZF (which = angle TZF – angle TZM) = angle AGN (which = angle AGD – angle NGD)].

2.147◉ Now, from point T draw a line TO parallel to line ZM, and this line will necessarily intersect FZ. Let the point of intersection be K. Take some line I that is to line ZT as [line] BG is to [line] EQ, [which is the] given line [i.e., I:ZT = BG:EQ, by Euclid, VI.12]. Then, at point M produce a conic section [i.e., a hyperbola] in the way Apollonius describes in the fourth proposition
of book 2 of his Conics, and let this conic section be UCM, which may not intersect lines KO and KF [that form its asymptotes].⁑ In this section draw a line equal to line I, i.e., MC, extend it to lines KT and KF, and let O and L be the
intersection-points [where MC extended cuts asymptotes KT and KF]. Thus, as the same [book of Apollonius’ Conics] will demonstrate, OM = CL.⁑

2.148◉ From point T draw line TF parallel to [line] CM, and at point A form an angle [GAN] with line AND that is equal to angle ZFT. It is evident that this line [AND] will intersect GD, since angle AGN = angle FZM, and angle GAN = angle ZFT. Hence, line AD will either be tangent to the circle or will intersect it, because, if it is not tangent, and if arc AB > arc AG, it will cut arc
AB, whereas if AB < [AG], it will cut arc AG.

2.150◉ But since TM is parallel to FL [by construction], and since FT is parallel to ML [by construction], FT = ML, so it will be equal to CO, since MO = LC [by Apollonius, II.8]. But MO = YT, since it is parallel to it, and YM is parallel to TO. It therefore follows that FY = CM. But CM = I [by construction]. Thus, FY = I. But I:ZT = BG:EQ [by construction]. Therefore, AN:GD = BG:EQ
[because we established earlier that AN:GD = FY:ZT].

2.152◉ But AD2 = BD,DG, as Euclid demonstrates [in III.36], and AD2 = AD,DN + AD,NA [by Euclid, II.2]. And BD,DG = DG2 + BG,GD, as Euclid demonstrates [in II.3]. Therefore, when equal terms are subtracted [i.e., GD2], it follows that AD,AN = BG,DG.⁑ Hence [by Euclid, VI.16], the second [AN] is to the fourth [DG] as the third [BG] is to the first [AD], so AN:DG = BG:AD. But it
has already been established that AN:GD = BG:EQ. Therefore, EQ = AD, which is what was proposed.

2.154◉ It is obvious that the two angles AHG and ABG sum up to two right angles [by Euclid, III.22]. But angle NGD = [alternate] angle ABG [because NG is parallel to AB, by construction]. Thus, angle AHG + angle NGD = two right angles. Accordingly, angle NGD = angle NHG [since angle AHG + angle NHG = two right angles], and angle NDG is common [to triangles NGD and HGD], so the third angle
[DNG] = the third angle [DGH], and so triangle HGD is similar to triangle NDG. Therefore, HD:DG = DG:DN, so HD,DN = GD2 [by Euclid, VI.17].

2.155◉ But AD,DH = BD,DG, as Euclid demonstrates,⁑ and AD,DH = DH,DN + DH,AN [by Euclid, II.1]. Moreover, BD,DG = BG,GD + GD2 [by Euclid, II.3]. Therefore, when equal terms are subtracted (i.e., GD2 and DH,DN), it follows that DH,AN = BG,DG,⁑ so [by Euclid, VI.16] the second term is to the fourth (i.e., AN:GD) as the third is to the first (i.e., BG:DH). But it has already been proven that AN:DG = BG:EQ. Thus, EQ = DH, and so what was proposed [has been demonstrated].

2.156◉ On the other hand, if AG < AB (and let [AD] cut [the circle] on arc AB), then let H [figure 5.2.19e, p. 232] be the point of intersection, and draw line HG. It is evident that angle NGD = angle ABG [by construction]. But [by Euclid, III.21] angles ABG and AHG are equal, since they are subtended by the same arc [AG]. Therefore, angle NGD = angle AHG, and angle NDG is common [to
triangles NGD and HGD], so the third angle [GND] = the third angle [HGD], and the triangles [NGD and HGD] are similar. Accordingly HD:GD = GD:DN, so HD,DN = GD2 [by Euclid, VI.17].

2.157◉ But HD,DA = BD,DG [since, by Euclid, III.36, both are equal to the square on the tangent drawn from D], and HD,DA = DN,HD + AN,HD [by Euclid, II.1]. Moreover BD,DG = DG2 + BG,DG [by Euclid, II.3]. Hence, with equal terms subtracted [i.e., GD2 and HD,DN], HD,NA = BG,DG. Accordingly, AN:DG = BG:HD [by Euclid, VI.16]. But it has already been established that AN:DG = BG:EQ. Therefore,
EQ = HD, which is what was proposed, because from point A we have drawn a line to cut the circle, and [the segment] from the point of intersection [H] to the diameter [BG] is equal to the given line.

2.158◉ [PROPOSITION 20, LEMMA 2] Now, from a given point on a circle outside its diameter, a line can be drawn through the diameter to the circle so that the segment on it between the diameter and the circle is equal to a given line.

2.159◉ For instance, let ABG [figure 5.2.20, p. 233] be the given circle, BG its diameter, A the given point, and HZ the given line. I say that a line can be drawn from point A to pass through diameter BG such that the segment [ED] from the diameter to the circle is equal to line HZ.

2.160◉ The proof [is as follows]. Draw lines AB and AG, and at point H form with line HM an angle [MHZ] equal to angle AGB, and at the same point form with line HL an angle [LHZ] equal to angle ABG. Then, from point Z draw a line ZN parallel to line HM, and it will intersect HL; and from point Z draw line ZT parallel to HL, and let it intersect HM at point T. At point T construct conic
section [i.e., hyperbola] TP, which Apollonius will describe in his book on conics [II.4], and that [conic] section will never touch either of the lines ZN and HL [i.e., the asymptotes] between which it lies. Between these same lines construct conic section [i.e., hyperbola] CU facing the first one.

2.161◉ Thus, if the shortest of all the lines extending from point T to [conic] section CU is equal to diameter BG, a circle produced according to this shortest line [as radius] when the point of a compass is placed at point T will be tangent to [conic] section CU. On the other hand, if the shortest of all the lines extending from point T to [conic] section CU is shorter than diameter BG,
the circle produced in the aforementioned way according to [radius] BG will intersect [conic] section CU at two points.⁑

2.162◉ Accordingly, let CT be the shortest [line], and [let it be] equal to diameter BG, and it will intersect ZQ and HF when it is extended to the [conic] section lying between them.⁑ From point Z draw a line parallel to this one, and it will intersect HM and HL just as its parallel [CT] does. Let it be MZL, let it intersect [those lines] at points M and L, let Q be the point where CT intersects ZN, and on diameter GB form an angle equal to angle HLZ, and let it be DGB. Then draw the two lines AD and BD.

2.163◉ It is evident that, since angle GAB is a right angle [by Euclid, III.31], the other two angles in triangle AGB sum up to a right angle, so angle LHM is right [since it consists of the two angles MHZ and ZHL that were constructed equal to angles AGB and ABG, which sum up to a right angle], and it is equal to angle GDB [which is a right angle by Euclid, III.31]. Also [by
construction], angle HLM [in triangle HLM] = angle DGB [in triangle DGB]. Hence, the third [angle HML is equal] to the third [angle DBG], and the [first] triangle [HML] is similar to the [second] triangle [GDB], so GB:BD = LM:MH.

2.164◉ But, since angle ADB = angle BGA, because they are subtended by the same arc [AB], and since angle BGA = angle MHZ, by construction, then angle ADB = angle MHZ. In addition, we already know that angle GBD = angle HMZ [by the similarity of triangles HML and DGB]. Hence, the third [angle in triangle DEB, i.e., angle DEB, is equal] to the third angle [in triangle MHZ, i.e., angle MZH],
so triangle DEB is similar to triangle MHZ. Let E be the point where line AD intersects diameter BG. Accordingly, BD:DE = MH:HZ. Hence, BG:DE = LM:HZ.⁑

2.165◉ However, Apollonius demonstrates [in Conics, II.16] that, when two [hyperbolic] conic sections lie opposite one another between two [asymptotic] lines, and when a line is drawn from one section to the other, the segment of that line lying between one of the sections and one of the [asymptotic] lines is equal to the opposite segment lying between the other section and the other
[asymptotic] line, so QC = TF. But TQ = MZ, since it is parallel to it [by construction] and lies between two parallels [HM and ZN, which are parallel by construction]. Therefore MZ = FC [since FC = TQ], and ZL = TF [since they are parallel and lie between TZ and HL, which are parallel by construction]. Hence, ML = TC, so GB:ED = TC:HZ, and since TC = BG, ED = HZ, which is what was proposed.

2.166◉ However, if the line extending from T to [conic] section CU is the shortest line and is shorter than diameter BG, then extend it beyond the [conic] section until it is equal. Then, according to its length [as radius], produce a circle that will intersect the [conic] section at two points, and from those points the lines extending to T will be equal to BG. Then, from point Z draw a
line parallel to both, and in that case two lines equal to the given lines will be drawn from point A in the prescribed manner, and this will be proven in precisely the same way.⁑

2.167◉ [PROPOSITION 21, LEMMA 3] Moreover, given a right triangle, and given some point on one of the sides forming the right angle, a line can be drawn from that point to the other side forming the right [angle] and intersecting the third side facing the right angle in such a way that the segment of this line that lies between the point of intersection and the side on which the given
point does not lie is to the segment of the side opposite the right angle from the [point of] intersection to the side containing the given point as a given line is to a given line.

2.168◉ For instance, ABG [figure 5.2.21, p. 235] is the given triangle with ABG the right angle, and the given point D, which is on side GB, is either in or outside the triangle. I say that from point D a line can be drawn to cut side AG and intersect side AB in such a way that the segment [TQ] on it between sides AB and AG is to the segment [TG] of the side AG extending from that line to
point G in the same proportion as E is to Z, those being the given lines [i.e., TQ:TG = E:Z].

2.169◉ The proof [is as follows]. Let point D lie on triangle ABG itself [as represented in figure 5.2.21], and draw from it a line DM that is parallel to AB. Produce a circle upon the three points G, M, and D [by Euclid, IV.5], and draw line AD. Since it is evident that angle GMD = [alternate] angle GAB, it will be greater than angle GAD. With line MN cut from [angle] GMD an angle equal
[to GAD], and let it be DMN, and [by Euclid, VI.12] draw line H so that AD is to it as E is to Z [i.e., AD:H = E:Z]. Then, from point N, which lies on the circle, draw a line to diameter GM⁑ that is equal to line H, according to the previous account
[in proposition 19, lemma 1], let it be [on line] NL, and let point C be where it intersects the circle [so that LC = H].⁑ Draw line GC, and from point D draw a line to point
C, a line which, when extended, necessarily intersects the other line [i.e., AB], since it falls between two parallels [AB and DM, which are parallel by construction] and forms with one of them an acute angle [i.e., MDT]. Let them intersect, then, and let Q be the point of intersection.

2.171◉ But angle NMD = angle TAD [i.e., GAD, by construction], and [it is also equal to] angle NCD [because they are both subtended by arc ND], so [angle] NCD [= vertical angle TCL in triangle TCL, and it also] equals [angle] TAD [in triangle TAD]. In addition, angle CTL is common to both triangles, so the third [angle TLC] = the third [angle TDA], and [so] the [one] triangle is similar to
the [other] triangle, i.e., [triangle] TLC to triangle TAD. Hence, TA:CT = AD:LC [but TA:CT = QT:TG, as established above], so AD:LC = QT:TG. But LC = line H, and AD:H = E:Z. Therefore, QT:TG = E:Z, which is what was proposed.

2.172◉ If, on the other hand, D is taken on that [same] side [that includes the right angle but extends] beyond the triangle [figures 5.2.21a and 5.2.21b, p. 235], then [in figure 5.2.21a] draw [a line] from point D that is parallel to AB, and let it be DM, and extend AG until it intersects DM at point M. Then produce a circle passing through the three points G, D, and M, and draw line AD.
Angle GAD [which is exterior to triangle DAM] will of course be greater than [interior] angle GMD. Form an angle equal to it [i.e., equal to GAD], and let it be NMD, and from point N, which is a point on the circle, draw a line equal to line H such that AD is to that line as E is to Z, and let that line be [on line] NCL, which is extended to diameter MG [so that CL = H and, therefore, that AD:CL = E:Z]. Let the point of intersection be L.

2.173◉ Thus, since angle NMD and angle NCD sum up to two right angles [by Euclid, III.22], and since angle NMD = angle TAD, the two triangles TCL and TAD will be similar.⁑ In addition, since the two angles GCD and GMD are equal [by Euclid, III.21], the two
triangles GCT and TAQ will be similar [so that DT:LT = TQ:TG = AD:CL], and so AD:CL (which = H) = QT:TG, and so E:Z = QT:TG, which is what was proposed.⁑

2.174◉ [PROPOSITION 22, LEMMA 4] Now, given two points, i.e., E and D, and given a circle, to find a point on it such that the line tangent to the circle at that point bisects the angle formed by the lines drawn from the aforementioned points to that point.

2.175◉ For example, from point E [figure 5.2.22, p. 236] draw line EG to the center of the given circle, extend it to the circumference, and let [this extended line] be ES. Then, draw line GD, and let line MI be cut at point C so that IC:CM = EG:GD [by Euclid, VI.10]. Then bisect MI at point N, and draw perpendicular NO. On point M form with line MO an angle [OMN] that is half of angle
DGS. It is clear that it will be smaller than a right angle, whereas angle ONM [will be] a right angle [by construction]. Thus, MO will intersect NO. Let it intersect at point O, and from point C draw a line CKF to the triangle [NMO] such that KF:FM = EG:GS [by proposition 21, lemma 3, case 2]. Then, with line AG, which is extended to the circle, form an angle at point G that is equal to angle MFK, and let it be angle AGE. Finally, draw the two lines AG and DG. I
say that A is the point we are seeking.

2.177◉ From point A, then, draw a line that forms with line AE an angle [EAZ] equal to angle NMK, and let it be line AZ, which will necessarily intersect line GE, because KF:FM = EG:GA, and angle GAZ = angle FMC [since angle EAG = angle FMK, and angle EAZ = angle KMN, by construction, so EAG + EAZ = GAZ = FMK + KMC = FMC]. Therefore, just as line MO will intersect [line] FK at point F, AZ
will intersect GE. Let the intersection occur at point Z, and draw AZ to point Q so that AZ:ZQ = MC:CI [by Euclid, VI.12], and draw line EQ.

2.180◉ Now, at point A form an angle equal to angle GAE, and let it be UAG. It is clear that angle GAL is half of angle UAT,⁑ but it is [also] half of angle DGU [since, by construction, OMN is half DGU, and GAL = OMN], so angle UAT = angle DGU. But angles TAU and TUA sum up to less than two right angles, since AT and UT intersect, so the two angles TUA and DGU sum up to less than two right angles. Therefore, AU will intersect DG.

2.181◉ I say that it will intersect at point D, because [by Euclid, VI.4] it will form a triangle with lines UG and GD that is similar to triangle AUT, for they will have angle AUG in common, and angle TAU = angle UGD [by previous demonstration]. Therefore AU:AT is as UG is to the line [X] that AU cuts from GD [i.e., AU:AT = UG:X], and EA:AU = EG:GU [by Euclid, VI.3], since angle UAG =
angle GAE [by construction].

2.182◉ Therefore, since EA:AT = EG:GD [by previous conclusions], EA:AT is compounded from EA:AU and AU:AT [i.e., EA:AT = (EA:AU):(AU:AT)].⁑ EG:GD will be compounded from these same ratios [i.e., EA:AU and AU:AT, so EG:GD = (EA:AU):(AU:AT), but AU:AT = UG:X, by previous conclusions] so it will be compounded from EG:GU and GU:X [i.e., EG:GD = (EG:GU):(GU:X)]. But it is compounded from EG:GU and GU:GD [i.e., EG:GD = (EG:GU):(GU:GD)]. Therefore the line that AU cuts off from GD is line GD [i.e., GD = X]. Therefore AU cuts GD at point D.

2.183◉ Accordingly, from point A extend tangent AH. GAH will therefore be a right angle. But [angle] GAL is half of angle DGU [by previous conclusions]. Hence, angle LAH is half of angle DGE, since these two [i.e., DGU and DGE] sum up to two right angles [so their halves, GAL and LAH, must sum up to a right angle—i.e., GAH, which is right by construction]. But since angle TAU = angle DGU,
angle TAD [adjacent to TAU] = [angle] DGE [adjacent to DGU]. Therefore, angle LAH is half of angle TAD [because LAH is half of DGE, by previous conclusions], and angle EAL is half of angle EAT [by previous conclusions]. Thus, angle EAH [which = EAL + LAH] is half of angle EAD, so AH bisects angle EAD, which is what was proposed.

2.184◉ On the other hand, if the angle at point A [i.e., UAG] = angle GAE, and if AU does not fall on line ES either outside or inside the circle, then let it be parallel [figure 5.2.22a, p. 236]. Accordingly, [alternate] angle UAG = [alternate] angle AGE. But the same angle [i.e., UAG] = angle GAE [by supposition], so angle GAE = angle AGE. Thus EG = AE. Likewise, angle TAD = angle ATG,
since they are alternate. But it has already been established that angle TAD = angle DGT [by previous conclusions]. Therefore, angle ATG = angle DGT, and by the same token the two angles ADG and DGT are equal. Therefore, the two angles ADG and TAD are equal [as are angles ADG and ATG, so triangle AMD is similar to triangle GMT, and both triangles are isosceles because of the equality of the angles at points A and D and at points G, and T].

2.185◉ From these conclusions it will therefore follow that the line AU cuts from DG is equal to line AT [i.e., GM = MT, and AM = MD, so AM + MT = GM + MD]. And it has already been established that EG = AE. Hence, EG is to the line AU cuts from DG [i.e., X] as AE is to AT [i.e., EG:X = AE:AT]. But it has already been established that AE:AT = EG:GD. Therefore the line that AU cuts off from
DG is GD, and since [angle] TAD = angle DGT, [angle] LAH will be half of angle TAD, as was claimed above, and [angle] EAL [will be] half of [angle] EAT [by previous conclusions]. Accordingly, [angle] EAH will be half of angle EAD, which is what was proposed.

2.186◉ [PROPOSITION 23, LEMMA 5] Moreover, given a circle with G as its center [figure 5.2.23, p. 237], given GB as a diameter within it, and given point E outside the circle, a line can be drawn from point E to diameter GB that cuts the circle in such a way that the segment of that line [extending] from the circle to the diameter [i.e., DZ] is equal to the segment of the diameter between
that line and the [circle’s] center [i.e., ZG].

2.187◉ For instance, from point E draw EC perpendicular to the diameter, and draw line EG. Take a line QT equal to EC, and on QT form a segment of a circle such that any angle within that segment [e.g., QPT] is equal to angle EGB [by Euclid, III.33], and then complete the circle. From the midpoint [L] of QT extend a perpendicular [FL] in both directions to the circle. This will of course
be a diameter of this circle. Then, from point Q draw a line to this diameter to intersect it at point F, and extend it to point P on the circle such that FP is half of GB [by proposition 20, lemma 2], and draw line PT and line TF. From point P draw line PU parallel to the diameter. Let it intersect TF at point U, and from point U draw UO parallel to TQ. From point T draw TN orthogonal to PQ, from point T draw TS parallel to PQ, and from point U draw UH orthogonal
to PQ. Then, from angle BGE cut off an angle BGD equal to angle QPU, and draw line EDZ. I say that DZ = ZG.

2.188◉ Now, from point D draw DI orthogonal to BG, and from point D draw tangent DK. It is clear that, since diameter FL is perpendicular to QT as well as to OU, and since PU is parallel to that diameter, angle OUP will be a right angle. And since OU is bisected by the diameter [FL] along the orthogonal, FO = FU, so angle FOU = angle FUO.⁑ However, since the [remaining] two angles POU and OPU [of triangle POU] sum up to a right angle, angle FUP = angle FPU [because triangle OFU = triangle PFU, since OU and OP are both bisected by FL, and OP is bisected
by FU], so FP = FU, and so it equals FO. And therefore PO = BG [because FP was constructed to be half of BG, and it forms half of PO], and it is also equal to GD [since GD and BG are both radii], so EC:GD = TQ:PO [since TQ = EC, by construction].

2.190◉ But QT:OU = TF:FU, since triangle TFQ is similar to triangle OFU [and QT and OU and TF and FU are corresponding sides]. However, angle UTS = angle HFU, since it is alternate to it [because PQ and ST are parallel, by construction], and right angle UST = [right] angle FHU [both being right by construction. Consequently] triangle UST will be similar to triangle HUF, so TU:UF = SU:UH,
and so [by Euclid, V.18] TF:UF = SH:UH. But TN = SH, since it is parallel to it and since both lie between two parallels [PQ and ST]. Therefore, TF:UF = TN:UH, so QT:OU [which = TF:UF, by previous conclusions] = TN:UH, and EC:DI [which = QT:OU, by previous conclusions] = TN:UH.

2.193◉ [PROPOSITION 24, LEMMA 6] Furthermore, given a right triangle ABG [figure 5.2.24, p. 238] with right angle ABG, and given point D on either BG or AB, to draw a line from point D to side AG intersecting it at point Q [on one side] and intersecting the other side [at point T] in the other direction such that the sum [of the lengths from D to the respective sides] is to GQ as E is to Z
[i.e., (TD + DQ = TQ):GQ = E:Z].

2.194◉ For instance, from point D draw DM parallel to AB, and produce a circle passing through the three points D, M, and G. MG will be a diameter [since angle MDG is right, by construction]. Draw line AD, and [by Euclid, VI.12] let H be a line in proportion to which AD is as E is to Z [i.e., AD:H = E:Z]. Since angle DMG = angle BAG, cut from it [an angle] equal to angle DAG, and let it be
CMD. Draw MC until it meets the circle at point C, and from that point draw a line to diameter MG and [extend it from point of intersection L] to the circle so that LN = line H [by proposition 20, lemma 2]. Then draw line NG, and [draw] line DN to intersect AG at point Q.

2.197◉ Furthermore [as demonstrated in proposition 20, lemma 2], it is possible for two lines like CN to be drawn, and in that case two lines can be drawn from D equal to TQ such that each of them is to the segment it cuts off from AG as E is to Z, and the proof will be identical.⁑

2.199◉ For instance, let A [figure 5.2.25, p. 239] be a center of sight, B a visible point, and G the center of the sphere [forming the mirror], and draw lines AG and BG. Take the plane within which these two lines lie, and take the [great] circle [that forms the] common [section] of this plane and the mirror. Accordingly, the point of reflection will be found on this circle.

2.200◉ Take some other line MK, and divide it at point F such that FM:FK = BG:GA [by Euclid, VI.10]. Bisect MK at point O, and from point O draw perpendicular CO, and from point K draw line KC to CO to form an angle [OCK] with CO that is equal to half of angle BGA. Then, from point F draw line FP to CK, and let it intersect CO at point S so that SP:PK = BG:radius GD [by proposition 24,
lemma 6]. From angle BGA cut off an angle equal to angle SPK, i.e., [angle] DGB, and draw lines SK and BD.

2.201◉ Accordingly, BG:GD = SP:PK [by construction], and so triangle SPK will be similar to triangle BGD [by Euclid, VI.6, because angle SPK = angle DGB, by construction], and [therefore] angle SKP = angle BDG. But, according to what we established earlier [in proposition 24, lemma 6], we may draw from point F to CK another line like SP [i.e., S’P’] such that it is to the segment it will
cut off from CK as SP is to PK [i.e., S’P’:P’K = SP:PK], and on that basis a line other than SK [i.e., S’K] will be drawn from point K to OS to form another angle [CKS’] with CK [i.e., other than the original CKS] that is either greater than, or less than angle CKS. If the larger of these angles is not greater than a right angle, no point of reflection will be found [as will be demonstrated below]. Accordingly, let angle CKS be greater than a right angle, and the
point [of reflection] is found as follows.

2.202◉ Angle BDG will be greater than a right angle [since BDG = CKS, by previous conclusions, and CKS is greater than a right angle, by stipulation]. Draw tangent NDY, and since angle PKO is smaller than a right angle, cut off angle QDG equal to it from angle BDG. Thus, since angle SPK = angle QGD [i.e., DGB, by construction], triangle FPK will be similar to triangle QGD [by Euclid, VI.4,
so angle DQG = angle KFP, and so] angle DQB [adjacent to DQG] = angle KFS [adjacent to KFP], and triangle DQB will be similar to triangle KFS [since angle SKP = angle BDG, and angle QDG = angle PKF, leaving remaining angles QBD and FKS equal].

2.203◉ Now, extend DQ [beyond Q], and from point B draw perpendicular BZ to it. Accordingly, angle BQZ = angle SFO [since triangles BQD and SKF are similar and BZ and SO are dropped orthogonally to corresponding sides from the corresponding vertices]. Also, right angle BZQ = [right] angle SOF, and so triangle BQZ is similar to triangle SFO [by Euclid, VI.4].

2.206◉ Now, from point D draw line DH to form an angle [HDL] with line LD that is equal to angle BGA. Then, since HL and DL intersect, the two angles LHD and LDH will sum up to less than two right angles, and so the two angles AGH [which = LDH, by construction] and DHG that are equal to these sum up to less than two right angles, so HD will intersect GA. I say that it will intersect at
point A.

2.207◉ It is evident that right [angle] GDN = the [sum of the] two angles OCK and OKC [in triangle OKC where KOC is a right angle, by construction, and angle NDG is also right, by construction], and angle OKC = angle GDQ [by construction]. It follows that angle QDN = angle OCK [because angles QDN and QDG sum up to a right angle, as do OCK and OKC, and angle QDG = angle OKC in similar
triangles QDG and FPK], so angle QDN = half of angle BGA as well as half of angle HDL [since angle QDN = angle OCK = half of angle BGA, by construction, and angle HDL = angle BGA, by construction]. But angle QDB is half of angle BDL, because BQ:QL = BD:DL, since triangle DLQ is similar to triangle BQI [by previous conclusions], and BD = BI [by previous conclusions]. It therefore follows that angle NDB is half of angle HDB, and so [angle] BDN = [angle] NDH. It
follows, moreover, that [when radius GD is extended beyond the circle to E, angle] BDE = angle HDG [because NDE and NDG are both right angles, and NDH = NDB, so NDE – NDB = BDE = NDG – NDH = HDG]. But angle HDG = vertical angle EDA, so [angle] BDE = [angle] EDA, and so D is the point of reflection. I say this is so if HD intersects AG at point A, which will be demonstrated as follows.

2.208◉ Draw line HT parallel to BD. It is clear that angle BDE = angle HDG [by previous conclusions]. But [angle] BDE = [alternate] angle HTD, so [triangle HDT is isosceles, and so] HT = HD. But [given that HT is parallel to base BD of triangle BDG, triangles BDG and HTG are similar, so], BD:HT = BG:GH, as Euclid demonstrates [in VI.2]. Thus, BD:DH [which = HT] = BG:GH. But, when it is
extended, HD will intersect GA, and it will form a triangle [HGA] similar to triangle HDL [by Euclid, VI.4], since it has angle LHD in common [with triangle HDL], and since angle HDL = angle HGA [by construction]. Therefore, HD is to DL as HG is to the line [X] that HD cuts from GA [i.e., HD:DL = HG:X]. But BD:DL is compounded from BD:DH and DH:DL [i.e., BD:DL = (BD:DH):(DH:DL)]. It is therefore compounded from BG:GH and GH:[X, i.e.,] the line HD cuts off from GA
[i.e., BD:DL = (BG:GH):(GH:X)].⁑ But BD:DL = BG:GA [by construction]. Therefore,
BG:GA is compounded from BG:GH and GH:[X, i.e.,] the line HD cuts from GA [i.e., BG:GA = (BG:GH):(GH:X)]. But it is compounded from BG:GH and GH:GA [so GA = X]. Therefore, GA is the line HD cuts off from GA, and so it will intersect it at point A, which is what was proposed.

2.210◉ For if it is claimed that [such reflection] can [occur], let D be the point of reflection, and draw line AD to point H on diameter BG. Make angle LDH equal to angle AGB, draw tangent NDY, and make angle QDN equal to half of angle AGB.

2.213◉ It is evident that BZ is perpendicular [to IQ]. Extend it until it intersects DG at point X, which is possible, because angle DZX is a right angle, while ZDX is less than a right angle. And it is obvious that BG:GD = SP:PK [by construction]. Therefore, since it is claimed that angle CKS is not greater than a right angle, I say that an angle greater than a right angle will be formed
at point K by a line that intersects CO at the point from which a line passing to CK through point F is to the segment of CK [cut off by it] as BG:GD [i.e., SP:PK = BG:GD].⁑

2.214◉ For example, it is clear that, since angle QDN = angle KCO [by previous conclusions], angle QDG = angle CKO. Accordingly, at point K form an angle equal to BDQ [i.e., SKO], and suppose that the line forming this angle intersects CO at point S, and draw SFP. It is evident that, since right angle BZD = [right] angle SOK, triangle BZD will be similar to [triangle] SOK, and BZ:BD =
OS:SK. But QZ:QD = OF:FK [by previous conclusions]. Hence, angle ZBQ = angle OSF, and angle QBD = angle FSK, so triangle BGD is similar to triangle SPK. Therefore, SP:PK = BG:GD, which is what was proposed.⁑

2.215◉ Furthermore, it is impossible for two angles to be erected on MO such that both of them are larger than a right angle. For if both such angles were larger than a right angle, then, since an angle equal to angle SKM may be formed upon the same center, another angle different from this one will be formed upon the same center which will form on KM another line like SK. And so
reflection will occur from point D as well as from some other point on this circle, which is impossible, since it has already been demonstrated [in proposition 16 above] that for any one center of sight there is [only] one point of reflection, and it has now been shown how to find it.⁑

2.216◉ Given two eyes, even though there are two points of reflection, there will nonetheless be a single image according to sense-deduction, and there will be a single image-location.⁑ We will
demonstrate this on the assumption that the two lines extending from the centers of the eyes to the center of the circle are equal.

2.217◉ Now, if the location of the visible point is the same with respect to both eyes, so that the lines [extending] from the visible point to the centers of the eyes are equal, the proof will be simple, because the visual axes cut an arc of reflection on the circle, and they form equal angles with the line extending from the visible point to the center of the sphere [i.e., the normal],
and the arcs lying between this line and the visual axes are equal. And if the points [of reflection] are selected according to the previous proof [in proposition 25 above], the arcs on the circle lying between these two points and the point on the circle that lies on the normal extended from the visible point [to the circle’s center] will be equal, which will be easily demonstrated by repeating the preceding demonstration.

2.218◉ And this is the case whether the points of reflection lie in the same plane of reflection, or in different ones; those arcs will still be equal, the lines extending from the centers of the eyes to the points of reflection will be equal, and the lines [extending] from the visible point to the same points will be equal. In addition, the lines extending from the centers of the eyes to
the points of reflection will necessarily intersect one another, and the proof is obvious that the intersection will be at the same point on the normal dropped from the visible point, and at this point one and the same image will appear to both eyes, which is what was proposed.⁑

2.219◉ Furthermore, the arrangement of images is the same as the arrangement of the visible points [producing them]. For if a line [forming a cross-section] is taken on a visible object, and if two lines are drawn from its endpoints to the center of the sphere, it will form a triangle within which the images of all the points on that [cross-sectional] line will be included. And if there is
a point on that line that is identically disposed [with respect to both eyes], the image of a point farther from it will lie on a normal farther from its normal, and one nearer [will appear] on a nearer [normal]. And so, in images [any] portion maintains its [relative] position as it actually exists among the visible points [within that portion].

2.220◉ Moreover, given a line on which there is [such] a uniformly disposed point, any point on that line will be uniformly disposed with respect to both eyes, according to the previously discussed way, and it will have a single image because of the equality of angles formed by that line with the visual axes. In addition, if a line is chosen that bisects the angle formed by the two lines
from the centers of the eye to the visible point, the location of any point on that line, no matter how far it is extended, will be the same for both eyes, just as was the case for the other line, and the same method of demonstration applies.

2.221◉ Aside from these two lines, no other one can be found that maintains the same location, so, when a visible point is perceived on the normal, its image will fall at different points on that normal, but [the two resulting images will lie] an imperceptible distance from each other. Also, the image of any point, no matter how many eyes may view it, always maintains a uniform relative
position, so the image appears unified, as has been claimed in the case of direct vision. For, even though they may fall at different places [on the normal], still, because of their insensible distance [from one another], the images do not appear disparate unless their relative positions are disparate. In a similar vein, when the distance of a point from one eye is slightly greater than it is from the other, the image-locations will [only] be imperceptibly
separated, so they appear fused, and one [image] is melded from them, in which case the image-locations may be partially rather than wholly distinct.⁑

2.222◉ In convex cylindrical mirrors, the common section of the plane of reflection and the mirror’s surface is sometimes a straight line, sometimes a circle, and sometimes a cylindric section [i.e., an ellipse].⁑

2.223◉ When the common section is a straight line, the image-location will lie on the normal extending from the visible point to the surface of the mirror, and that image-location lies just as far [below the mirror] from the common section [i.e., the line of longitude] as the visible point lies [above it]. [This is subject to] the same proof as applies to the plane mirror [in proposition 2
above].

2.224◉ However, when the common section is a circle, the image-location will sometimes lie inside the circle, sometimes outside it, and sometimes on the circle itself. This phenomenon has the same explanation as applies in the convex spherical mirror [in propositions 11-15 above].

2.225◉ But if the common section is a cylindric section, I say that some of the image-locations lie inside the mirror, some on the mirror’s surface, and some outside the mirror. These claims will be explained individually.

2.226◉ [PROPOSITION 26] Let ABG [figure 5.2.26, p. 241] be a cylindric section, B the point of reflection, E the visible point, and D the center of sight. From point B draw a normal to the plane tangent to the mirror at point B, let it be TBQ, and from point E draw perpendicular EKQ to the plane tangent to the mirror at point K [to form the normal dropped from the object-point]. Let the
line tangent to the mirror at point B be CU; let the line tangent to the mirror at point K be KM. I say that the two normals TB and EQ will intersect.

2.227◉ Draw lines EB and DB, and draw line KB. It is obvious that KM will fall within figure EKB, and line BC [will fall] within the same figure. Therefore, BC will intersect EK. Let it intersect at point C. It is evident that angle TBK > right angle [CBT], and angle EKB is likewise greater than right angle [EKM], so TB and EK will intersect [since angles BKQ and KBQ, adjacent to angles
EKB and TBK are acute]. Let Q be the point of intersection. Likewise, [angle] DBK is greater than a right angle. Therefore, DB and EK will intersect. Let H be the point of intersection. Accordingly, H is the image-location. I say, as well, that EQ:QH = EC:CH, and also that QH > HB.

2.230◉ From this it is evident that, if a perpendicular is dropped to a plane tangent to segment GB [of the mirror], it will intersect TB. By the same token, any [perpendicular] dropped to segment AB [of the mirror] will intersect TB. And these conclusions are evident when the visible point does not lie on the visual axis. For [when it does] it is clear from earlier discussions [in
proposition 8 above] that the form of one single point reaches the mirror orthogonally and is reflected back along the same line, and this point [whose form reaches the mirror along the] normal lies on the surface of the eye, for a point taken outside the eye cannot be reflected along this normal because it cannot reach the mirror along this normal according to the aforementioned reasoning [i.e., that the body of the eye gets in the way]. By the same token, it
could not be reflected from a point on the mirror other than from a point on the normal, because [in that case] two normals would happen to intersect and form a triangle with two right angles, just as was shown above [in proposition 8].

2.231◉ [PROPOSITION 27] Furthermore, take [some] cylindric section [figure 5.2.27, p. 242], select point A on it, draw tangent [E]AT to the section [at point A], and within the [section on the] mirror take DA perpendicular to [E]AT.

2.232◉ It is obvious that AD divides the [cylindric] section into two portions, in either of which there is a single point to which the tangent will be parallel to AD. Accordingly, let G [be a point in one of the two portions of the section] to which the tangent intersects AD at point H, and to this tangent draw perpendicular QG, which will necessarily intersect HD, as was shown in the
preceding figure [i.e., 5.2.26]. Let D be the point of intersection, draw line GA to P, and draw line QA. Accordingly, angle QAH is equal to, greater than, or smaller than angle HAP.

2.233◉ Let it be equal. The form of point Q will therefore reach A and will be reflected to P, which is the center of sight, and the image-location will be a point on the cylindric section, i.e., G [because G is where line of reflection PA intersects normal QD].

2.235◉ But if some point, such as C, is taken between Q and T, angle CAH > angle HAP. Make it equal to HAM. It is clear that [line of reflection] MA will fall on [normal] GQ, but outside the [cylindric] section. Let it [do so] at point O. Thus, the image of point C will lie at point O, and the images of all points lying between T and Q will lie outside the [cylindric] section between T
and G.

2.236◉ Moreover, if angle QAH < angle HAP, then cut [an angle] from HAP equal to it, and let it be HAN. It is evident that the image of Q will lie at point K, and the images of all points above [Q] will lie within the [cylindric] section. But if point C is taken below [Q] so that angle CAH = angle HAP, the image of C will lie on the [cylindric] section, [the images of] all [points]
between C and Q [will lie] inside [the cylinder], and [the images of] all [points] between C and T [will lie] outside [the cylinder].

2.237◉ If, however, angle QAH > angle HAP, make [angle] HAM [figure 5.2.27a, p. 242] equal to it. It is clear that MA will intersect the [cylindric] section [at some point beyond A on arc AG]. Let it intersect at point B, draw the tangent to point B, and let it intersect DH at point L. Now, angle DLB will be acute, angle HLB will be obtuse, and, when it intersects HG, LB will form an
acute [angle] with it. Draw SB perpendicular to LB at point B. It will intersect HG, it will form an acute angle with it, and its vertical angle will likewise be acute. Let HG intersect QA. Let U be the point of intersection, and it forms an acute angle with QA at point U [i.e., angle HUA is acute], so SB and QU intersect. Let the intersection be at point Z. It is therefore evident that the form of point Z will reach the mirror along ZA, and it is reflected along
AM [since QAH = HAM, by construction], and B [will be] the image-location. Moreover, the images of points above Z on line ZS will lie inside the [cylindric] section [because both the new ZAH and MAH will be more acute], whereas [the images] of points below Z [on line ZS] will lie outside the [cylindric] section [because both the new ZAH and the new MAH will be more obtuse], which was what was proposed.⁑

2.238◉ [PROPOSITION 28] Moreover, reflection occurs to a center of sight from only one point on a cylindrical mirror, as, for example, [the form of] point B [figure 5.2.28, p. 243] is reflected to [point] A from point G. I say that it does not reflect to the same point from any point on the mirror other than from point G.

2.239◉ For if the entire axis [CED] of the mirror lies in the plane of reflection ABG, the common section of the mirror’s surface and the plane of reflection will be a line of longitude [FGN] on the mirror. And since the center of sight [A], the visible point [B], the point of reflection [G], and the point [E] on the axis where the normal [to the point of reflection] falls [all] lie in the
[same] plane of reflection, only one plane can be assumed within which that line of longitude, or the axis, and points A and B lie, so reflection can only occur to A from some point on the line of longitude [FN]. But it has already been demonstrated [in proposition 3] that reflection cannot occur to [point] A from any point other than G on the line of longitude, so in this case reflection occurs to A from only one point on the mirror.⁑

2.240◉ But if the plane [of reflection] A’B’G is parallel to the base of the cylinder, the common section [of this plane with the cylinder] will be a circle [GH] parallel to the base. And it has already been demonstrated [in proposition 16] that reflection to [point] A’ cannot occur from any other point on that circle. But if reflection were to occur from some other point on the mirror
[outside circle GH], the normal dropped from that point would fall orthogonally to the axis [CED], and it would intersect line A’B’ at some point [K]. From that point draw a line [KLI] to the axis in the plane parallel to the base of the cylinder. It will of course be orthogonal to the axis, and so two perpendiculars [KE and KI] will form with the axis a triangle [KEI] two of whose angles [KEI and KIE] are right angles, which is impossible. It is therefore evident
that in this case B does not reflect to A except from point G.⁑

2.242◉ From [the center of sight at] point A [figure 5.2.28a, p. 243] produce a plane parallel to the base of the cylinder, let it be EZI, and likewise from point [of reflection] G produce a plane parallel to the base of the mirror [i.e., GSP] and in that plane draw line TG from the [mirror’s] axis [TQ] to point G. This line will therefore be perpendicular to the plane tangent to the
mirror at point G. Let it intersect [line] AB at point K, and from point G draw GZ, the line of longitude on the mirror, and let TQ be the axis. Then, from [object-]point B draw BH perpendicular to plane EZI, and draw lines AZ and HZ. In that same plane [i.e., EZI] draw line ZQ from Z to the axis. It will be perpendicular to the axis, since the axis is perpendicular to this plane [EZI within which it lies], and it will be perpendicular to the plane tangent to the
mirror at point Z. Let it also intersect line AH at point L. I say that the form of point H is reflected to A from point Z.

2.243◉ From point A draw [line] AM parallel to line KG, and it will intersect BG. Let M be the point of intersection. It is evident that [line] GZ is parallel to line BH, since [by construction] both of them are perpendicular to parallel planes [i.e., EZI and GSP], so line BGM lies in the same plane [BGMH] as these lines [GZ and BH]. Accordingly, the three points M, Z, and H lie in this
plane. But AM in turn is parallel to KG, and LZ is parallel to KG, because GZ is parallel to TQ and lies between parallel planes [i.e., EZI and GSP]. Therefore, LZ is parallel to AM, so they lie in the same plane [AHZM], and line AH lies in it too. Hence, the three points M, Z, and H lie in this [same] plane. But it has already been shown that they lie in plane M[G]BH. So these two planes [AHZM and MGBH] intersect along a common line [MZH]. Therefore, HZM is a
straight line.

2.244◉ Since G is the point of reflection, it is evident that angle [of reflection] AGK = angle [of incidence] KGB, and so angle AGK = angle AMG [which is alternate to angle KGB between parallels AM and KG]. But it is equal to MAG, because it is alternate to it [between the same two parallels]. Hence, [within isosceles triangle MAG] AG and MG are equal. But since GZ is orthogonal to any
line within plane AZH, [then, by the Pythagorean Theorem] MG2 = MZ2 + GZ2 [in right triangle MZG]. By the same token, AG2 = AZ2 + GZ2 [in right triangle AZG]. Hence, AZ = MZ [because AG = MG, by previous conclusions], so angle AMZ = angle ZAM. But [since AM and LZ are parallel, by previous conclusions] angle AMZ is also equal to [alternate] angle LZH, and angle ZAM = angle LZA, since they are alternate angles. Therefore, angle AZL = angle LZH, so, when it reaches
point Z, the form of point H is reflected to point A.

2.245◉ Hence, if it is claimed that the form of B can be reflected to A from some point [D] other than G, that other point will lie either on line of longitude GZ or on some other [line of longitude]. If it lies on GZ, draw from it a perpendicular [DL’], which will necessarily intersect line AK and will be parallel to line AM [because, by supposition, it must lie in a plane of reflection
that includes points A and B, and it must be normal to the mirror]. In addition, the line extending from point B to that point [D] will necessarily intersect line AM [because line AM is in the plane of reflection that includes A and B], and [so] that point and point M will lie in the same plane.

2.246◉ Furthermore, that line [BD] will either fall on point M or [will fall] on some other point. If [it falls] on point M, then two [different] straight lines will have been drawn from point B to point M [i.e., BGM and BDM]. On the other hand, if [it falls] to another point on line AM [i.e., N], draw a line from that point to point Z, and it is demonstrated that this line forms a
straight line with HZ, as was demonstrated for line ZM. And so two straight lines will have been drawn from point H to pass through point Z and fall at different points on AM, which is impossible.

2.247◉ It is therefore evident that [the form of] B can be reflected to A from no point other than G on line GZ. If it is claimed that [it can be so reflected] from a point taken outside this line, draw the line of longitude on the mirror that [lies] on this point, and from the point on circle EZI where this line [of longitude] falls, it is demonstrated that [the form of] H is reflected to
A according to the previous proof. But it has already been demonstrated that [the form of] H is reflected to A from point Z, so [the foregoing conclusion] is impossible [by proposition 16 above]. It therefore follows that [the form of] B is reflected to A from only one point on the mirror, which is what was proposed.

2.248◉ [PROPOSITION 29] Furthermore, given that [the form of] point B is reflected to A, it will be possible to find the point of reflection, and this will be demonstrated by reversing the [previous] proof.

2.249◉ From point A [figure 5.2.28a, p. 243] produce a plane parallel to the cylinder’s base, and this plane will cut the cylinder along circle EZI. Draw line BH orthogonal to this plane from point B, and find point Z within this plane from which the reflection of [the form of point] H occurs to A [by proposition 25 above]. From point Z draw line of longitude ZG, and from point Z [draw]
perpendicular ZL, to which AM [extended] from point A is parallel. Extend line HZ until it intersects the latter, and let M be the point of intersection. From point M draw a [straight] line to B, and it will necessarily intersect line ZG, since it lies in the same plane with it. Consequently, since BH is parallel to GZ, HZM will lie in the same plane with them, and so MB [will lie] in that same plane, and if it intersects ZG at point G, then point G will be the
point of reflection, and you can see this if you reverse the previous demonstration.⁑

2.250◉ In convex conical mirrors, if the common section of the plane of reflection and the surface of the mirror is a line of longitude on the mirror, the image-location will be just as it was determined in the case of plane mirrors, and the same proof [applies].

2.251◉ That the common section cannot be a circle is evident from the fact that the normal [must] fall orthogonally to the plane tangent to the mirror at the point of reflection, and [the plane of] the circle will necessarily be parallel to the [mirror’s] base. But no [plane] parallel to [the plane of] the base will be orthogonal to a plane tangent to the mirror.

2.252◉ If, however, the common section is a conic section, some images will lie on the surface of the mirror, some inside the mirror, and some outside it. And [all this] is determined by the same method as [was applied] in the case of the convex cylindrical mirror, and the same proof [applies]. Furthermore, just as was shown in the case of the convex cylindrical [mirror], the form of only
one point on the surface of the eye is reflected orthogonally along the visual axis to the eye, and it does so from only one point on the mirror, and its image-location is continuous with the other image-locations [surrounding it], as was shown above.

2.254◉ [PROPOSITION 30] Let A [figure 5.2.30, p. 244] be the center of sight, B the visible point, G the point of reflection, and on point G produce a plane parallel to the base, a plane that will cut the cone along circle PG. Draw lines AG, BG, and AB, and from point G draw line GT to the center of the circle. Let the vertex [of the cone] be E, and from it draw axis ET. Then [in the plane
of ETG] draw perpendicular HG to the plane [containing line CG that is] tangent to the mirror at point G, and, since [normal] HG bisects angle AGB, it will fall upon AB. Let Z be the point where it falls [on AB].

2.255◉ From the vertex draw line of longitude EG on the mirror to point G, and draw a line from point A parallel to this line, and this parallel will necessarily intersect the plane of circle GP. Let it be AN, and let it intersect at point N. Likewise, from point B draw a parallel to EG, i.e., BM, and let it intersect the plane [of circle] PG at point M. From point N, as well, draw NF
parallel to GT, and draw lines NG, MG, and NM.

2.256◉ It is evident that TG will intersect NM. Let it intersect at point Q. It is also evident that MG will intersect NF, since it intersects [line TG] parallel to it. Let F be the point of intersection. Then, from point A draw AL parallel to HZ. It is clear that BG will intersect AL [since A, B, G, and Z lie in the same plane, and AL is parallel to HZ, by construction]. Let L be the
[point of] intersection. Then draw GC, which is the common section of the plane [CGE] tangent to the mirror at point G and the plane of circle PG. It is clear that it will be orthogonal to GT, as well as to NF [because NF was constructed parallel to GT].

2.257◉ In addition, take GD, which is the common section of the plane tangent [to the mirror at point G] and the plane of reflection [AGBZ], and GD will intersect AL, since it intersects GH [to which AL was constructed parallel]. Let D be the point of intersection, and GD will be perpendicular to AL [since it is perpendicular to ZG, which is perpendicular to plane CGED tangent to the
mirror at point G and parallel to AL].

2.258◉ It is clear from the foregoing that NF is parallel to GT [by construction], and AL is parallel to GH [by construction]. Therefore, the plane containing NF and AL [i.e., FLAN] is parallel to plane G[E]TH. But EG is parallel to BM, so they lie in the same plane, and that plane intersects the aforementioned parallel [planes, i.e., GETH and FLAN], one along line [of longitude] EG, the
other along line FL, so FL is parallel to EG. But AN is parallel to that same line [EG, by construction]. Therefore, FL is parallel to AN.

2.259◉ However, the plane tangent to the mirror at point G intersects the same parallel planes [i.e., GETH and FLAN], one along line [of longitude] EG, the other along line CD. Therefore, CD is parallel to EG. It is therefore parallel to AN and LF, so AD:DL = NC:CF [by Euclid, VI.1, VI.2, and V.11].

2.261◉ I say, then, that [the form of] point B is reflected to A only from G. For if it is claimed that it can be reflected from some other point, that point will either lie on line of longitude EG, or it will not.

2.262◉ Let it lie on it, let it be X [figure 5.2.30a, p. 245], and from that point draw the normal to the plane tangent to the mirror at that point, and this normal will be parallel to ZG and therefore parallel to AL. AL will therefore lie in the plane of reflection containing this normal, and it will likewise lie in the plane of reflection containing normal ZG. Hence, those two planes of
reflection intersect along line AL. But they intersect on point B, which is impossible, because B does not lie on line AL, which is evident from the fact that FL is parallel to BM.⁑ It therefore follows that [the form of point] B can be reflected to A from no other point on line EG than G.

2.263◉ But if [it were assumed to do so] from some other point [not on line of longitude EG], then let that point be U, draw line of longitude EUO, and take the plane parallel to base [PG of the mirror] passing through point U. It is clear that AN will intersect this plane. Let Y be the point of intersection. BM will intersect the same plane as well. Let K be the point of intersection, and
draw lines KU, YU, and YK. Since that plane intersects the cone along a circle passing through U, draw line RU from point U to the center of this circle. Draw lines EK and EY, which [when extended] will intersect the plane of circle PG, and let I and S be the points of intersection. Now, draw lines IO and SO.

2.264◉ Thus, just as it has been demonstrated [earlier in this proposition] for point M that, barring interference from the cone, it[s form] can be reflected to N from point G, so it is demonstrable for point K that it[s form] can be reflected to point Y from point U, and the proof is the same. And so angle RUY = angle RUK.

2.265◉ It is therefore evident that BK is parallel to EG [by construction], and the common section of plane BGEK and the plane of circle PG is line MG. Therefore, since it lies in that plane [i.e., BGEK] and intersects the plane of circle PG, line EK will fall on common section MG. SMG will therefore be a straight line.

2.266◉ By the same token, since plane NYEG intersects the plane of circle PG along line NG, line EY will intersect line NG [at point I]. Thus, ING is a straight line. It is also evident that plane IOE intersects the plane of circle PG along line IO, and it intersects the plane parallel to this one that passes through U along line YU. Therefore, YU is parallel to IO. Likewise, plane SOEK
intersects those parallel planes along the two lines SO and KU. Therefore, SO is parallel to KU.

2.267◉ Similarly, if the plane cutting the mirror along line of longitude EO and containing R, U, O, and M is taken, it will intersect those parallel planes along the two lines MO and RU. Therefore, these two lines are parallel. Hence, angle SOM = angle KUR, and angle MOI = angle RUY. But it has already been shown that angle KUR = [angle] RUY. Accordingly, angle SOM = angle MOI, so point S
can be reflected to I from point O, barring interference from the cone.

2.268◉ But it has already been shown [earlier in this proposition] that [the form of] point M can be reflected to I from point G, and so [the form of] point S, which lies on [straight] line SMG, can be reflected to I from point G. Therefore, [the form of] point S is reflected to I from two points on circle PG, which is impossible. It follows, then, that the main [supposition of this
proposition], i.e., that [the form of] point B can be reflected to A from some point other than G on the mirror, is impossible, which is what was proposed.

2.270◉ For instance, Let G [figure 5.2.31, p. 246] be the vertex of the cone, and at that point produce plane MNG parallel to the base of the cone. Let A be the visible point and B the center of sight. A and B will [both] lie above that plane [i.e., MNG]; or [they will both lie] below it; or [they will both lie] in the plane itself; or one [will lie] above it [while] the other [lies] below
it; or one [will lie] in the plane [while] the other [lies] above or below it.

2.271◉ Let them lie below the plane, and from point A produce a plane that cuts the cone parallel to the base, and from point G to point B draw a line, which, when extended, will fall on the plane produced from A [through the cone], since it lies between parallel planes [i.e., MGN and HEA]. Let H be the point where this line falls [on that plane].

2.272◉ Now, it is proven according to the previous method [applied in proposition 30 above] that [the form of] A is reflected to H from some point on the circle formed on the cone by the plane [HAT] produced from points A and H. Let the point of reflection be found on that circle [by proposition 25], and let it be E. Then draw line AB, and draw the cone’s line of longitude GE, as well as
axis GT of the cone.

2.273◉ From point E draw line ET to the center of the circle, and it will fall to the axis, and it will be orthogonal to the plane tangent to that circle at point E. Then, with lines AE and HE drawn in, it will bisect the angle formed by them, and it will divide line AH [in half]. Let R be the point of division.

2.274◉ It is evident that GE and ET form a plane that cuts line AB. Let F be the point of intersection, and from point F draw normal FC to line GE, and it will be perpendicular to the plane tangent to the cone along line GE. Then, from point A draw AL parallel to line FC. FC, moreover, will intersect the axis at point K. From point A draw AS parallel to line RT, and from point E draw
common section EO of plane AEH and the plane tangent to the cone along GE. It will fall orthogonally to AS, since it is orthogonal to ER [to which AS is parallel, by construction].

2.275◉ Draw line BC, which, when extended, necessarily intersects line AL [which is parallel to FC, by construction]. Let the intersection be at point L, and from point C draw common section CP of the plane tangent [to the cone along GE] and plane ABL. Draw lines LS and PO.

2.276◉ It is obvious that plane ALS is parallel to plane GEK [because AS is parallel to RT, by construction, AL is parallel to FC, by construction, and FC and RT lie in the same plane], and lines CE and PO lie in the plane tangent [to the cone along GE], that plane intersecting those parallel planes [ALS and GEK] along the two lines CE and PO. Hence, CE is parallel to PO.

2.277◉ Furthermore, draw line HE until it intersects AS at point S. It is clear that line ES lies in plane HEG, and BL lies in the same [plane], and this plane cuts the aforementioned parallel planes [ALS and GEK] along the two lines EC and LS. Therefore EC is parallel to LS. Therefore PO will be parallel to LS [since it is parallel to CE], so AO:OS = AP:PL [by Euclid, VI.2].

2.279◉ But if the center of sight and the visible point [both] lie in plane MGN [figure 5.2.31a, p. 247], let the former be at point M, the latter at point N, draw lines MG, NG, and MN, and bisect [angle] MGN with line UG. It is clear that [the form of] N is reflected from G to M.⁑ It is also clear that line UG and the cone’s axis lie in a plane that cuts the cone along a line of longitude [GE].

2.280◉ From point U draw UE perpendicular to this line of longitude. Through point E produce the plane parallel to the base [of the cone], and this plane will cut the cone along a circle. Let ET be the common section of plane UEG and this circle. It is evident that it will fall on the axis as well as on the center of the circle.

2.281◉ Then, from point M draw a line [MH] parallel to GE, a line that falls at point H on the plane of that circle. Likewise, from point N draw a line [NA] parallel to GE and falling at point A [on the plane of the circle]. Draw AH, and let ET intersect it at point R.

2.282◉ It is obvious that MH, which is parallel to GE, lies in the same plane with it, and this plane [MHEG] intersects plane MGN and plane HEA along the two lines MG and HE. Therefore [because planes MGN and HEA are parallel, by construction], MG is parallel to HE. By the same token, AN and GE lie in the plane [ANGE] that cuts those parallel planes [MGN and HEA] along NG and AE. Hence, NG
is parallel to AE. Likewise, plane UGE intersects the same planes [MGN and HEA] along the two lines RE and UG. Thus, UG and MG are parallel to HE and RE [in reverse order], so angle MGU = angle HER, angle UGN = angle REA, and angle HER = angle REA. And so [the form of] point A can be reflected to H from point E.

2.283◉ Accordingly, if a line is drawn from point A parallel to UE, if another [is drawn] parallel to RE, if ME is extended until it intersects the line parallel to UE, if the common sections are drawn, as before, and if the preceding proof is repeated, it will be clear that [the form of point] N can be reflected to M from point E.⁑ E will therefore be the point of reflection, which is what was proposed.

2.284◉ Now, if both [the center of sight and the visible point] lie above MGN [figure 5.2.31b, p. 247], construct the cone opposite the original one. To do so, extend the lines of longitude of the previously constructed cone, and through point A pass a plane that cuts this latter cone parallel to the base, and it will cut the cone along circle YZ.

2.285◉ B will either lie in this plane, or it will not. If it does, then carry out the procedure from point B. If not, then extend line GB until it intersects this plane. Let the intersection be at point D. It is evident that A is reflected to D from some point inside circle YZ.⁑ Find that point (as we will later prove and explain [in proposition 38 below], it is not among those on the anterior [surface of the cone]), and let it be Z. Lines DZ and AZ will be drawn, and let line
PZ bisect the angle [formed by them].

2.286◉ Extend line ZG to the other cone, and it will reach its surface to form a line of longitude [on it]. Let it be line ZGE. It is obvious that plane PZE will intersect line AB. Let it intersect at point Q, and from point Q draw a perpendicular to line GE, and let it fall at point E.⁑ It will also be perpendicular to the plane tangent to the cone along line GE. Through point E produce plane A’EH parallel to the base [of the cone], and from point D draw line DH parallel to ZE and intersecting that plane at point H. Let A’A be parallel to that same line.

2.287◉ It is clear that DH is parallel to ZE, and they lie in the same plane [DHEZ], which intersects the parallel planes [A’EH and AZD] along the two lines DZ and HE. Accordingly, HE and DZ are parallel. Likewise, AZ and A’E are parallel. And it is evident that PZ passes through the center of circle YZ, and so does RET [pass] through the center of the other circle along which plane A’EH
cuts the cone. Therefore, plane PZER intersects the two parallel planes [AZD and A’EH] along the two lines PZ and RE. PZ, then, is parallel to RE, so angle AZP = angle A’ER. And therefore angle A’ER = angle REH, so [the form of] A’ is reflected to H from point E.

2.288◉ Accordingly, if a line is drawn from point A’ parallel to QE, if another [line is drawn] parallel to RE, and if the common sections [are drawn] as before, and if we repeat the previous method of proof, it will be evident that [the form of] point A is reflected to B from point E, which is what was proposed.⁑

2.289◉ If, however, the center of sight lies in the plane parallel to the [cone’s] base and passing through its vertex, i.e., [point] G, and if the visible point lies below this plane, the point of reflection will be found in the following way.

2.290◉ Let M [figure 5.2.31c, p. 248] be the center of sight, A the visible point, and MGN the plane parallel to the base of the cone. From point A produce a plane parallel to the base of the cone, and it will cut the cone along circle DEK with centerpoint T. From point M draw MH perpendicular to this plane, and draw line HT. From point A draw line AEQ to HT within the circle such that EQ
= QT, according to the earlier account [in proposition 23, lemma 5]. Then, draw line TEI, and from point H draw HB parallel and equal to TE. Draw lines MB and BE. It is evident that plane GTE will intersect line AM. Let F be the point of intersection, and from point F draw FOC perpendicular to line GE and intersecting it at point O. Then, draw lines MO and AO. I say that O is the point of reflection.

2.291◉ It is clear that HB is parallel and equal to TE [by construction]. Therefore, HT is parallel and equal to BE [since they are cut by equal and parallel lines HB and TE so as to form a parallelogram]. But MH is parallel and equal to GT, since both are perpendicular [to parallel planes DEK and MGN]. Accordingly, HT is parallel and equal to MG. Therefore, MG is parallel and equal to BE
[since BE is parallel and equal to HT, by previous conclusions], so MB is parallel and equal to GE.

2.292◉ It is also clear that angle QTE = angle QET [because QT = EQ, by construction, so triangle EQT is isosceles], and so it is equal to angle AEI [which is the vertical angle of QET]. But it is [also] equal to angle IEB [which is alternate to QTE]. Thus, [angle] IEB = angle IEA, so [the form of] A is reflected to B from point E. And since MB is parallel to GE, if a line parallel to FOC,
as well as to TE, is drawn from point A, and if the earlier figure and proof [based on figure 5.2.31, p. 282] are repeated, it is clear that A is reflected to M from point O, and so [we have done] what was proposed.

2.293◉ Now, if M lies in the plane [of MGN], and A lies above that plane, then the cone opposite the original one will be produced. Pass a plane through A that [cuts the opposite cone] parallel to its base, and the point of reflection is found among points lying inside the circle formed by this plane. Draw the line to G from that point, and extend it. And the point of reflection will be
found according to subsequent analysis [in proposition 38 below], and the same method of proof will apply.⁑

2.294◉ But if the [relevant] points, i.e., the center of sight and the visible point, are so disposed that one of them lies above the plane of the [cone’s] vertex and the other below it, let one of them be L [figure 5.2.31e, p. 249] and the other A, and let the plane at the vertex be MGN.

2.295◉ Through point A extend a plane parallel to the base of the cone and intersecting it along circle DE with center T, and draw line LG. It will intersect plane AED. Let K be the [point of] intersection, and in circle DE find point E such that tangent SE drawn from that point bisects the angle formed by lines KE and AE [by proposition 22, lemma 4].

2.296◉ Then, from point L draw a line [LB] parallel to [line of longitude] GE, a line that will necessarily intersect line KE. Let B be the intersection. It is clear that L lies in plane GEK, and LB, being parallel to GE, lies in that same plane. Draw line TEI. It is evident that plane GTE intersects line LA. Let it intersect at point U, and from there draw normal UOC to the plane tangent
[to the cone at point O]. Finally, draw lines AO and LO.

2.297◉ It is clear that [angle] AES = angle SEK [by construction], and since angle IES is a right angle, and [angle] SET is a right angle, [angle] IEA [which = IES + AES] = angle TEK [which = SET + SEK]. And therefore angle AEI = angle IEB [from which it follows that angle of incidence AET = angle of reflection TEB], so [the form of] A is reflected to B from point E. If, then, a line
parallel to UO and a line parallel to IT are drawn from point A, and if the proof [from the previous cases] is repeated, it will be clear that [the form of] A is reflected from point O to L, and so [we have done] what was proposed.⁑

2.298◉ It is therefore clear how the point of reflection can be found, and what has been discussed must be understood to apply to a single center of sight. But in [the case of] both eyes, the same thing happens, because the same form and the same location for the form is perceived by each eye, and, as has been claimed in the case of the convex spherical mirror, the forms perceived by both
eyes in these sorts of mirror [i.e., conical convex] appear identical because of their proximity; sometimes they share precisely the same [image-]location, sometimes their [image-]locations overlap, and sometimes they are separated, but only a little bit.

2.300◉ [PROPOSITION 32] In the case of spherical concave mirrors, the normal dropped from the visible point [to the mirror] sometimes intersects the line of reflection, and sometimes it is parallel to it. When it intersects, the image-location will sometimes lie on the [surface of the] mirror, sometimes behind the mirror, and sometimes in front of it. And when the image-location lies in
front of the mirror, it will sometimes lie between the center of sight and the mirror, sometimes at the center of sight [itself], and sometimes beyond the center of sight. And we will demonstrate this [as follows].

2.301◉ Let A [figure 5.2.32, p. 250] be the center of sight and D the center of the mirror, and through these points produce a plane that will cut the mirror along [great] circle HBFG. This plane will be a plane of reflection, because it is orthogonal to any plane that is tangent to the [great] circle [on the sphere]. Draw line AD, and from point A to the circle draw line AE, which is
longer than AD. From point D to the circle draw [line] DH parallel to line AE, extend AD to points B and I [on the circle], and draw line DE.

2.302◉ It is evident that angle AED is less than a right angle, because ED is a radius, and [by Euclid, III.31] any line in a circle forms an acute angle with the radius [or any other segment of the diameter]. At point E form angle DET equal to angle AED. It is clear that ET will fall inside the circle [since DET is acute] and will intersect line DH. Let T be the point of intersection. It
is also clear that angle ADE > angle DET, so ET will intersect AB. Let it intersect at point Z.

2.303◉ Then, from point A draw line AN to arc EH, draw line DN, and at point N form with line NM an angle [DNM] equal to angle DNA, and this line [NM] will necessarily fall inside the circle [because it cannot be tangent at N] and will intersect DH. Let it intersect at point M. It is clear, as well, that AN will intersect DH outside the circle [since DAN and ADH sum up to less than two
right angles]. Let L be the intersection.

2.304◉ In addition, draw line AG from point A to arc EF, draw DG, and let angle AGD = angle DGQ. It is obvious that QG will intersect DH. Let Q be the point of intersection. It is also obvious that AG will intersect DH on the side of F. Let O be the intersection. Moreover, since the arc [GY] that subtends GO within the circle is greater than arc GH, it is evident that GQ falls between D
and H. For if line GH is drawn, angle HGD[P] will be subtended by a greater arc [i.e., HP] than angle [P]DGA[Y, i.e., arc PY].

2.305◉ Furthermore, from point A draw line AC to arc FB so as to cut DH at point S in such a way that CS > SD, and draw DC. It is clear that angle DCA is acute. Form angle DCK equal to it. Since angle CDS > angle DCS, it is evident that CK will intersect DH. Let K be the point of intersection.

2.306◉ According to the foregoing construction [making angle DET = angle AED], it is clear that [the form of] point T propagates to E and is reflected to A. TD is the normal dropped from point T, and this line, which is perpendicular to the plane tangent to the circle [at H], is [by construction] parallel to AE, the line of reflection, so it will not intersect it.

2.307◉ [The form of] point Z, on the other hand, propagates to E and is reflected to A. AZ is the normal dropped from point Z, and it intersects AE at point A, so the image-location for point Z will be [at center of sight] A.

2.308◉ On the other hand, [the form of] point M propagates to N and is reflected to A. Normal MD dropped from point M intersects [line of reflection] AN at point L, which lies behind the mirror, and the image-location for point M will be L.

2.311◉ From the foregoing it is thus clear that some of the images lie behind the mirror, some between the center of sight and the mirror, some at the center of sight itself, and some beyond the center of sight, which is what was proposed.

2.312◉ It is evident, moreover, that the visual faculty grasps forms that face it, so, when the image-location lies behind the mirror or between the eye and the mirror, the image is grasped according to how it is actually located. However, when the normal dropped from the visible point is parallel to the line of reflection, the image will appear at the point of reflection. For, since that
[visible] point is [a] sensible [spot] represented by the point imagined [to lie] at its center, the image of any portion of that sensible spot taken beyond the midpoint [toward the mirror] will lie behind the mirror, whereas the image of the portion in front of the midpoint [away from the mirror] will lie between the eye and the mirror, and since the whole form appears as a continuum composed of the parts behind and the parts in front [of the mirror], the form of
that sensible spot will necessarily appear in an image-location on the mirror itself.⁑

2.313◉ But in the case of images that are located at the center of sight, they are not grasped according to how they are actually located, so [visual] deception often occurs in these sorts of mirrors [i.e., spherical concave]. To make this clear, stand a wooden rod less than half the radius of the mirror in length upright upon the surface of the mirror [along a normal]. Let the center of
sight be situated directly above this rod, and look at a spot on the mirror that lies farther from the rod than the center of sight does [from the mirror] along the normal passing through the rod. The image of this rod will appear behind the object itself, but it will not be perceived properly; on the contrary, it will appear bowed when it is not.⁑ Hence, in these sorts of mirrors the image is seen according to how it is actually located only when the image-location lies behind the mirror or between the eye and the mirror. But when the center of sight lies on the normal passing through the rod, it does not perceive the form of that rod clearly.

2.314◉ On the other hand, if the center of sight is located on a diameter of the sphere and at its center, then, since every line dropped from it to the mirror is perpendicular to the mirror, the form of no point will be perceived except a point within the portion of the circle lying between the edges of the visual cone that is imagined to extend [to the mirror’s surface] from the center
of the circle [where the center of sight is located]. For the form of any other point will fall on the mirror along an oblique line, and it is necessarily reflected along an oblique line, so the line of reflection will not pass through the center, and so it will not reach the center of sight.⁑

2.315◉ However, if the center of sight lies on a diameter but not at the center, it will not perceive the form of any point on the radius within which it lies. For the angle that the two lines from the given point on the radius to the center of sight will form at the same point on the mirror will not be bisected by the normal extended from that point on the mirror, since that normal is
directed to the center of the mirror. But it can perceive the form of any point on [a] radius other [than the one opposite it on the diameter upon which it lies].⁑

2.316◉ [PROPOSITION 33] Now, when a point is viewed in this sort of mirror, if the normal is not parallel to the line of reflection, the line extended from the center of the mirror to the visible point will have to the line extended from that same centerpoint to the image-location the same ratio that the line drawn from the visible point to the point we have called the [end]point of
tangency has to the line extended from the [end]point of tangency to the image-location.

2.317◉ For example, let E [figure 5.2.33, p. 251] be the center of the mirror, B the visible point, A the center of sight, G the point of reflection, and ZG the line of tangency. Either ZG will intersect EB, or it will be parallel to it.

2.318◉ Let it intersect at point T. Line EB will intersect AG, but not at point G, since EB and BG are two [distinct] lines. Therefore, they will intersect behind G, or between G and A, or at A, or in front of A. Let [the intersection be] behind G, at point H. Accordingly, I say that EB:EH = BT:TH.⁑

2.319◉ Draw normal EG, and from point H draw [line HL] parallel to line BG, so it will intersect EG. Let L be the intersection, and from point B draw [line BQ] parallel to GH, so it will necessarily intersect ZT. Let Q be the intersection.

2.321◉ But since angle GHT = [alternate] angle TBQ, triangle TBQ will be similar to triangle GHT [because they have two equal corresponding angles, i.e., angles TBQ and GHT and angle BTQ and vertical angle GTH]. Thus, QB:HG = BT:TH [by Euclid, VI.4], and so BG:HL = BT:TH [because we have established that BG:HL = QB:HG, and QB:HG = BT:TH]. But since triangle BGE is similar to triangle HEL
[HL and GB being parallel, by construction], then BG:HL = EB:EH, and so EB:EH = BT:TH, which is what was proposed.

2.323◉ If, however, line of tangency ZG is parallel to normal BH [figure 5.2.33d, p. 252], draw perpendicular GE, which, since it is perpendicular to ZG, will be perpendicular to BH. And [right] angle BEG = [right] angle HEG, and angle BGE = angle EGH. It follows that triangle BGE is similar to triangle EGH. Therefore, BE:EH = BG:GH, which is what was proposed, because in this case, no
[end]point of tangency other than G can be assumed, according to the way we defined the [end]point of tangency earlier [in proposition 6, where the endpoint of tangency lies at the intersection of the normal dropped from the object-point and the line tangent to the point of reflection].

2.324◉ [PROPOSITION 34] Now, let DGT [figure 5.2.34, p. 253] be the [great] circle [cut from the mirror by the plane of reflection], A the center of sight inside the circle, E the center of the mirror, and B the visible point. Draw diameter DAG.

2.325◉ If B lies on radius EG, there can be reflection from some point on semicircle GTD as well as from some point on the semicircle opposite [and below] it. For if a line is drawn to some point [F] on semicircle GTD from some selected point [B] on radius EG, and if another line is drawn from point A to the same point [F on semicircle GTD], those two lines will form an angle that the
radius [EF] extended from point E will bisect at that point. The same holds in the case of the opposite semicircle.

2.326◉ But if B lies outside diameter DAG [figure 5.2.34a, p. 253], draw the diameter passing through B, and let it be TBQ. I say that [the form of] B can be reflected to A from the arc lying between the radii in which A and B lie, and likewise from the arc opposite it, i.e., from arc TD and from arc GQ, but it cannot be reflected from any point on arc GT or arc QD.

2.327◉ For instance, take some [potential] point [of reflection] K on arc GT, and draw lines AK and KB until KB intersects diameter DG at point O. Since O and A lie on the same side of E, which is the center of the circle, the normal dropped from point K to E will not divide angle OKA, and so [the form of] B is not reflected to A from point K. Likewise, if another [potential] point [of
reflection] F is chosen [on that same arc], it will be obvious that normal EF will not divide angle AFB, and so [the form of] B is not reflected to A from point F.

2.328◉ That reflection can, however, take place from a point on arc TD or on arc GQ is clear from the following. Let M be a [potential] point [of reflection] on arc DT, and draw lines AM and MB to form quadrilateral AMBE. Therefore, normal EM will divide angle AMB.

2.329◉ By the same token, let H be a [potential] point [of reflection] on arc GQ. Line AH will intersect diameter TQ at point C, and line HB [will intersect] the same [diameter] at point B. And these two points lie on different sides of the centerpoint [E], so line EH will divide that angle.

2.330◉ Likewise, if B lies on the surface of the mirror, or outside the mirror, as long as A lies inside the mirror, the same method of proof will apply as before. And the same holds if A lies on the surface of the mirror, and B lies inside or outside it.⁑

2.331◉ But if AP is drawn from A parallel to TE [figure 5.2.34b, p. 254], the locations of images reflected from points [such as M] on arc TP will lie outside [and behind] the mirror, whereas the locations of images [reflected from points such as N] within arc PD will lie [at points, such as S] beyond the center of sight A, and the locations of images [reflected from points such as H]
within arc QG lie [at points, such as X] between the center of sight and the mirror. And the same thing that has been just said about image-locations must be understood [to apply] when AM is drawn parallel to line TQ.

2.332◉ On the other hand, if A lies outside the mirror and B inside it [figure 5.2.34c, p. 254], what we claimed will be evident [as follows]. From point A draw lines AH and AZ tangent to circle GTD, draw the two diameters AEG and TEQ, and let B lie on diameter TEQ. [The form of] B is reflected to A from some point on arc TD, but it is obvious that [it is] not [reflected] from any point on
arc ZD. Therefore, [it is reflected] from some point on arc TZ, and likewise from some point on arc GQ opposite TD. However, according to the previous method [of demonstration], reflection will not occur from arc TG or [arc] DQ.

2.333◉ If, however, B lies outside this diameter [TEQ] and on another, which is likewise [designated] T’EQ’, reflection will occur from arc T’D, but only within portion T’Z on it, and [it will occur] from GQ’, the arc opposite it. Reflection will not occur, however, from arc T’G or [arc] DQ’.

2.334◉ If, however, some point, such as point F, is taken above point Q, angle FAH < angle HAP. Make it equal to [angle] NAH. [Line of reflection] NA will intersect [normal] GQ inside the cylinder. Let it [do so] at point K. It is evident, then, that the image of point F will lie at point K, and the images of all points beyond Q [will lie] inside the cylinder.

2.334 [PROPOSITION 35] Furthermore, if a diameter is taken on a [great] circle within a spherical concave mirror, any point on that diameter, no matter how far it is extended, can be an image-location.

2.335◉ For instance, let AG [figure 5.2.35, p. 255] be a diameter of circle AMG, whose center is D. Take point Z on this diameter, [and take] E [as] the center of sight. I say that Z can be an image-location.

2.336◉ For instance, draw line ETZ, T being a point on the circle. Draw line DT. Angle ETD will be acute. Form [angle] DTL equal to it. It is clear [from the equality of angles ETD and DTL] that [the form of] L is reflected to E from point T and that its image will be Z.

2.338◉ Now, among the points that are perceived in these sorts of mirrors, some of their images are distributed in four locations, some in three, some in two, and some in [only] one. A point whose image lies at four locations is reflected from four distinct points, and from no others nor from more [than four]. A point whose image occupies three locations is reflected from three points on
the mirror, and from no more than three; one whose [image occupies] two [locations is reflected] from two points [only]; whereas one whose image lies in a single location may have its reflection occur from one point only, and it may [have it occur] from some point on a specific [great] circle, but from no other [point within the mirror].

2.339◉ [PROPOSITION 36] For example, let E [figure 5.2.36, p. 255] be the center of sight, let H be a visible point on the same diameter, and let D be the center of the circle. Draw diameter ZEHA. ED is either equal to DH, or not.

2.340◉ Let it be equal, and from point D on EH draw diameter GDB perpendicular [to EH], and draw lines HG, GE, HB, and BE. It is evident that triangle HGD = triangle EGD, and it equals triangle HBD, as well as triangle EBD. Since angle HGE is bisected [by diameter GDB], it is clear that [the form of point] H is reflected from point G to E, and its image-location is E [where line of
reflection GE intersects normal HD dropped from visible point H]. Likewise, [the form of point] H is reflected to E from point B, and its image-location is E.

2.341◉ Therefore, if diameter ZEHA remains stationary and semicircle AGZ, or only triangle HGE, is rotated [around AZ, as an axis] throughout [the circumference of] the sphere, point G will describe a circle in the course of its motion, and from any point on that circle [the form of point] H is reflected to [point] E, and its image-location will always be point E, and so [we have
demonstrated] what was proposed.

2.342◉ That reflection of [the form of] point H to E cannot take place from any point other than one on that circle [of revolution] is evident from the following. Take point C. Line EC > line EG, and line HC < line HG, so EC:HC ≠ ED:DH [which means that angle ECD ≠ angle DCH, by Euclid, VI.3]. Therefore, line DC will not bisect angle ECH, so [the form of point] H cannot be reflected
to E from point C. The same disproof will apply if C is taken between G and Z.

2.344◉ Produce a circle with a radius of QD having its center at Q, let G and B be the points where the two circles intersect, and draw lines EG, EB, QG, QB, DG, DB, HG, and HB. Accordingly, it is evident that EQ:QG = QG:QH [because radius QD = radius QG, and, as established earlier, EQ:QD = QD:QH], and angle GQH is common to both triangles EQG and HQG [whereas angles QHG and QGE are
right]. Therefore, those two triangles are similar. So [by Euclid, VI.4] EQ:QG = EG:GH. Therefore, ED:DH = EG:GH [since we established that EQ:QG = EQ:QD = ED:DH], so line DG will bisect angle EGH [by Euclid, VI.3].

2.345◉ Consequently, [the form of] point H is reflected to E from point G, and its image-location is point E. Likewise, [the form of point] H is reflected from point B to E, and its image-location is E.

2.346◉ Hence, if points E and H are held stationary and triangle EHG is rotated [around QE, as axis], point G will describe within the sphere a circle from any point of which [the form of] H is reflected to E, and E will always be the image-location.

2.347◉ Moreover, as before, it is clear that [the form of point] H cannot be reflected to point E from any point [within the sphere] other than one on that circle. For if C is taken between G and A, EC > EG, and HC < HG, so EC:HC ≠ ED:DH, and so [by Euclid, VI.3] DC does not bisect angle ECH. Likewise, if C is taken between G and Z, [then the supposition that reflection can occur
from C] can be disproved.

2.348◉ And so what was set out [has been demonstrated], as long as it is borne in mind that E is an imaginary point, and the circle [of revolution] with E as pole is an imaginary circle, and H is an imaginary point. Therefore, what has been claimed according to geometrical demonstration is not to be understood according to ocular proof, because imaginary objects are invisible to sight. But
since the form of H appears continuous with the forms of other points [in close proximity within the visible spot represented by H], the form that sight will see is a form whose midpoint is H, and the midpoint of that form [when reflected to E] will be E, and this form will be reflected from a circular segment of the mirror within which the aforementioned circle [created by the rotation of G] will represent the midline with E as its pole.

2.349◉ Moreover, if ED > DH, insofar as it can be enough longer that [the form of point] H does not reflect to E from point G, it must be understood that, unless EA:AH > ED:DH, [the form of] H cannot be reflected to E.

2.350◉ For if it could be reflected, let it be reflected from point G. Angle GDH will be less than a right angle, since it is subtended by an arc less than a quarter of the circle. From point G draw the tangent, which will necessarily intersect EA. Let Q’ be the intersection. By the [33d proposition⁑ we know that] EQ’:Q’H = ED:DH, but EA:AH > EQ’:Q’H. Therefore, EA:AH > ED:DH, and so, by necessity, if [the form of point] H is reflected to E, EA:AH > ED:DH.⁑ The claims that have been made are therefore obvious when the center of sight and the visible point lie on the same diameter.

2.351◉ [PROPOSITION 37] Furthermore, if the visible point and the center of sight do not lie on the same diameter, and if they lie outside the mirror, the [form of the] visible point is reflected to the center of sight from only one point on the mirror.

2.352◉ For instance, let T [figure 5.2.37, p. 257] be the visible point, H the center of sight, and D the center of the sphere, and draw lines HD and TD. Plane HDT intersects the sphere along circle EBQG.

2.353◉ It is clear that [the form of] T is not reflected to H except from some point on this circle. According to earlier discussion [in proposition 34 above], it is also clear that it is not reflected from arc QG or BA. It is therefore reflected from either arc GB or AQ.

2.354◉ Bisect angle TDH with line LEDZ, and draw tangent KEF from point E. If points T and H lie on that tangent, [the form of point] T will not be reflected to H from any point on arc BG. For if a line will be drawn from point T to some point on this arc [on the] inner [side of the circle], the line drawn from point H to the same point will fall to it on the outer side [of the circle],
not the inner side, and so there will be no reflection.⁑

2.356◉ Either lines TD and HD are equal, or they are not equal. If they are equal, and if line DZ is common, then triangle TDZ = triangle HZD, and angle TZH is bisected by line DZ, and so [the form of point] T is reflected to H from point Z.

2.357◉ That it cannot [be reflected] from another point [on arc AQ] will be established as follows. Take point O, draw lines TO and HO, and let line ODM bisect the angle [formed by them]. It is clear [by Euclid, III.8] that TZ < TO and that HO < HZ, and [by Euclid, VI.3 it is also clear that] TZ:HZ = TL:LH [since angle HTZ is bisected by ZL, leaving triangles HZL and TZL equal] and
TO:HO = TM:MH [since angle TOH is supposedly bisected by OM, leaving triangles HOM and TOM equal]. But HO:TO < HZ:TZ [since TO > HO, and HZ = TZ]. Accordingly, HM:MT < HL:LT, which is impossible [because, according to the bisection of angle HOT by HM, HM:MT = HO:TO, whereas, according to the bisection of angle HZT by ZL, HZ:TZ = HL:HT, so it should necessarily follow that HM:MT = HL:LT, because M and L should cut equiproportional segments from HT].

2.358◉ It is therefore obvious that, if T and H lie the same distance from the center [of the mirror], and if they lie above the tangent, [the form of point] T is reflected to H from only one point on the mirror, and it will have a single image-location.⁑

2.359◉ Now, let BDQ and ADG [figure 5.2.37b, p. 258] be two diameters within the sphere [from which the mirror is formed], let diameter EDZ bisect angle BDG, and from point E draw the two [lines] ET and EH perpendicular to the two diameters BD and GD.

2.360◉ It is obvious that triangle ETD = triangle EHD, since ED is common to both [and angles EHD and ETD, as well as EDH and ETD are equal, by construction]. Thus, [the form of point] T is reflected to H from point E. By the same token, [it is reflected to point T] from point Z. It is also obvious [from proposition 34 above] that it is not reflected to [point] E from any point on arc AB
or arc GQ, nor is it reflected from any point on arc AQ other than from point Z, according to the previous demonstration [in case 1 above]. But that it cannot be reflected from any point on arc BG other than from point E will be shown as follows.

2.361◉ Given point O [from which reflection supposedly does occur], draw lines TO, HO, and DO. Construct a circle the size of DE [as diameter] that passes through the three points T, D, and H, and line DE will be the diameter [by Euclid, III.31], since the angle ETD that it subtends is a right angle [by construction]. Therefore, that circle will pass through point E.

2.362◉ Accordingly, since E is common to each circle [i.e., TDHE and AZGE], and since it lies on the same diameter [within each circle], the smaller circle will be tangent to the larger circle at point E, as Euclid demonstrates [in III.11]. Therefore, this [smaller] circle will intersect line DO.

2.363◉ Let it intersect at point I, and draw lines TI and HI. It is already evident that TD = DH [since they are corresponding sides of equal triangles TED and HED]. Thus, angle TID = angle DIH, since they are subtended by equal arcs [DT and DH in circle HTE]. It follows that [being adjacent to angle TID] angle TIO = angle HIO [which is adjacent to angle DIH]; angle IOT = angle IOH, by
construction, and IO is common. [Hence] triangle TIO = triangle HIO [by Euclid, I.26], and TO = HO, which is impossible, since [by Euclid, III.7] HO > HE, [while] TO < TE, and TE = HE, as has been proven earlier. It therefore follows that [the form of point] T may not be reflected to H from any point other than E or Z.

2.364◉ Moreover, from point E draw line EM to diameter TD, from line HD cut off a segment ND equal to MD, and draw EN and EM. It is evident that angle EMD is greater than a right angle [because it is exterior to right angle ETM in triangle ETM]. Cut from it [an angle] equal to a right angle with line CM, which will intersect DE. Let C be the point of intersection, draw NC, and construct a
circle according to the size [of diameter] CD that passes through the three points M, D, and N. Since CMD is a right angle [by construction], CD will be the diameter, and the circle will pass through C. It is therefore obvious that [the form of point] M is reflected to N from point E [because triangle NCE = triangle MCE, so angle NED = corresponding angle MED], and likewise [it is reflected] from point Z, but from no other point on arc AB or QG; and it is obvious
that [it does not reflect] from any point on arc AQ other than point Z.

2.365◉ That [it does not reflect] from any point on arc BG other than from point E is evident from the earlier procedure. For if [some such] point is taken, if lines are drawn to that point from points T, D, and H, if a point is taken where the last circle will cut the diameter, and if lines are drawn to points T and H from the point of intersection, the same disproof as before will
apply.

2.366◉ From the foregoing, then, it is clear that, if a third diameter bisects the angle formed by two [other] diameters, and if from the endpoint of that [third] diameter perpendiculars are drawn to those [other two] diameters, the points on the diameters where those perpendiculars fall are reflected to one another from only two points on the mirror. In addition, any of the points taken
on the perpendiculars to the diameters below these endpoints, i.e., toward the center, is reflected from only two points [on the circle], and the one is reflected to the other that lies equidistant from the center, and for all such points the image-location is twofold.⁑

2.367◉ Furthermore, given the two diameters BQ and AG [figure 5.2.37c, p. 259], and given that EZ bisects the angle formed by them, take point T on BD beyond the point [between D and T] where the perpendicular dropped from point E falls [upon DB], take DH on DG equal to DT, and draw TE and HE. [Thus, according to the preceding case, the form of point] T is reflected to H from point E, and
likewise from point Z, [but] not from any other point on arc AQ, or from any point on arcs AB or GQ.

2.368◉ Then, from point T on TD draw a perpendicular, which will intersect DE outside the circle on the [mirror’s defining] sphere, because angle DTE is acute. Accordingly, let it intersect at point O, and draw lines TO and HO. Construct a circle passing through the three points T, D, and H. That circle will necessarily pass through point O, and DO will be its diameter. Draw lines TO and
HO, and draw line KE tangent to circle BZG at point E. It is clear that the latter circle will intersect the first one, i.e., BZG, at two points. Let those points be L and M, and draw lines TL, HL, LD, TM, DM, and HM.

2.370◉ But [the form of point] T cannot be reflected to H from any point other than these. For, let F be given [as such a point], draw lines TF, HF, and DF, extend DF until it intersects tangent KE, and let K be the intersection. Then draw lines TK and HK. Hence, angle TFD = angle DFH, by construction; and it follows that [being adjacent to angle TFD] angle TFK = angle KFH [which is
adjacent to angle DFH]. But angle TKF[D] = angle [D]FKH, because they are subtended by equal arcs, and FK is common. [Therefore] triangle [TKF] = triangle [HFK], and so TK = KH, which is impossible, because HK > HO [since arc HK > arc HO, whereas] TK < TO, yet TO = HO [by construction].

2.372◉ [In summary] therefore, if two points, i.e., T and H, are taken on different diameters [at locations on those diameters that are] equidistant from the center, and if they lie at the points on the diameters where the perpendiculars drawn from the end of the diameter that bisects the angle formed by those two diameters fall, or if they lie between the center and those points, i.e.,
below the perpendiculars, as long as they lie equidistant from the center, then [the form of point] T will be reflected to H from two points only [as demonstrated in case 2 above].

2.373◉ If, however, T and H lie at spots beyond the perpendiculars up to the interior surface of the circle, [the form of point] T will be reflected to [point] H from four points. But if they lie on the circle or outside it, but still below tangent KE, [the form of point] T will be reflected to H from two points only [i.e., E and Z]. And if they lie upon the tangent, [the form of point] T
will be reflected to H from only one point [i.e., Z].⁑ And these phenomena occur as long as T lies the same distance as point H from the center.

2.375◉ For instance, draw diameters ADG and BDQ’ [figure 5.2.38, p. 260], and let EZ bisect the angle formed by them. Let [object-point] T be nearer centerpoint D than [center of sight] H. Take [some arbitrary] line LQ and cut it at point M so that QM:ML = HD:DT [by Euclid, VI.10]. Bisect LQ at point N, from point N draw perpendicular NK, and at point L form with line FL an angle [FLQ]
equal to half of angle ADT. Angle FLQ will be acute, so FL will intersect NK. Let it intersect at point F, and from point M draw a line to side FL intersecting side NK at point K. Let that line cut side FL at point C such that KC:CL = HD:DZ [by proposition 21, lemma 3 above].

2.376◉ Then, on point D form angle IDA equal to angle LCM, and let I be a point on the circle lying above or below Z. At point I form angle O’ID equal to angle CLM, to the resulting line O’I drop perpendicular HC’ from point H, draw line C’F’ equal to line C’I, and draw lines HF’ and HI.

2.377◉ According to previous arguments, it is clear that from point M no line other than line MCK can be drawn to side FL to cut it in the way that line MCK does. For if [such a line] could [be found], let it be MPO. It is evident that PO < CK, which will become evident if line PY is drawn parallel to CK, that line being shorter than CK and longer than PO. But PL > CL. Thus, PO:PL ≠
KC:CL, so PO:PL ≠ HD:DT. It therefore follows that no line like MCK, other than MCK [itself], can be drawn from point M [to side FL].

2.380◉ Now, if line UI is drawn from point I parallel to HF’, and if line DA is extended until it intersects UI at point U, triangle O’UI will be similar to triangle HO’F’. Therefore, HO’:O’U = QM:ML [because HO’:O’U = F’O’:O’I], and so HO’:O’U = HD:DT [because QM:ML = HD:DT, by construction]. But since triangle HC’I = triangle HC’F’, because HC’ is perpendicular [to equal bases F’C’ and
C’I], then angle HF’C’ = angle C’IH, so [angle] C’IH = angle UIO’, and so [by Euclid, VI.3] HO’:O’U = HI:UI [because angle UIH in triangle UIH is bisected by IO’, insofar as angle UHI is composed of equal angles UIO’ and C’IH], and so HI:UI = HD:DT.

2.382◉ It is evident that the ratio HI:UI is compounded from HI:IP and IP:UI, while HI:IP = HD:DP [by Euclid, VI.3], because DI bisects angle PIH [in triangle PIH]. Therefore, HI:UI, which is as HD:DP, is compounded of the ratios HD:DP [which = HI:IP] and PI:UI. But the ratio HD:DT is compounded of HD:DP and DP:DT. Thus, DP:DT = PI:UI.⁑

2.383◉ Now, angle O’IH is half of angle UIH [by previous conclusions], whereas angle DIH is half of angle PIH [by construction]. It follows that angle DIO’ is half of angle PIU, whereas angle DIO’ is half of angle TDP, since it is equal to angle FLM. Therefore, angle PIU = angle TDP,⁑ and DP:DT = PI:UI [by previous conclusions]. Hence, triangle UIP is similar to triangle TPD [by Euclid, VI.4], and [so] angle UPI = [angle] TPD. TPI will thus be a straight line, because angle DPT + angle TPO’ = two right angles, and so angle O’PI + angle O’PT = two right angles [since O’PI is alternate to DPT]. So [the form of point] T is reflected to [point] H from point I, and this proof will apply whether T lies outside or inside the circle, and
[it will apply] likewise if point H is taken outside or inside [the circle], as long as [they are] unequally distant from the center.⁑

2.384◉ [PROPOSITION 39] Now, having produced diameters BQ and AG, and [given] diameter EZ bisecting angle BDG, I say that, no matter what point other than Z is taken on arc AQ, an infinite number of point-pairs that are not equidistant from the center can be reflected [to one another] from that point.

2.385◉ For example, take point H [as the potential point of reflection in figure 5.2.39, p. 261], and take point L on diameter GD. On diameter BD cut off [segment] MD equal to LD, and draw lines LM, LH, MH, and DH. Let F be the point where EZ cuts LM; [and so] LF = FM [since triangles FMD and LDF are equal, by Euclid, I.4].

2.386◉ Draw [line] HD until it falls on LM at point N. Hence, LN < MN. But, since angle MDF = [angle] FDL, as well as [vertical] angle QDZ, since angle MDA = [vertical] angle LDQ, and since angle ADH = [vertical] angle NDL, then angle LDH > angle MDH.⁑ Therefore [by Euclid, I.24], LH > MH, since MD and DH = LD and DH [respectively]. Thus, angle DHL < angle DHM, for if they
were equal [then DHN would bisect angle MHD, and so, by Euclid, VI.3], LH:MH would be as LN:NM, which is impossible. If, on the other hand, it were greater [i.e., if DHL > DHM], then cut from it [an angle] equal [to DHM], and it will be disproven in this [same] way.⁑ Therefore it is smaller [i.e., DHL < DHM].

2.388◉ So, too, if points other than L and M that lie equidistant from point D are selected on diameters HD and GD, it is proven in a similar way that from point H there occurs a mutual reflection of points that lie at different distances from the center. And so the proof will be the same for an infinite number of points chosen on these diameters, as well as for any point chosen on arc AQ
other than point Z [because, in order for reflection to occur from Z, the points must be equidistant from the center, as shown in proposition 37].⁑

2.389◉ [PROPOSITION 40] Moreover, if T and L [figure 5.2.40, p. 261] are taken on [different] diameters at unequal distances from the center, and if they are reflected to one another from point H, [the form of point] T will not be reflected to [point] L from any point on arc AQ other than from point H.

2.390◉ For if [it were reflected] from another [point], let it be K, and draw TK, LK, DK, LT, TH, LH, and NDH. Then extend DK until it falls to point C on LT. It is clear that LH:TH = LN:NT [by Euclid, VI.3].

2.391◉ Likewise, since angle TKC = angle LKC, by supposition, then LK:TK = LC:CT. But [by Euclid, III.7] LH > LK, and TH < TK. Hence LH:TH > LK:TK, so LN:NT > LC:CT, which is patently impossible. It follows that [the form of point] T cannot be reflected to L from any point on arc AQ other than from point H. What pertains to arc AQ is therefore evident.

2.392◉ [PROPOSITION 41] To continue, let A [figure 5.2.41, p. 262] be the center of sight, B the center of the mirror, and draw diameter DABG. Take the plane in which AB lies so that it will somehow intersect the sphere along [great] circle DLG. I say that points that do not lie the same distance from the center as A are reflected [to A] from any point on semicircle DLG.

2.393◉ For instance, take point E, and draw lines EA and EB. It is obvious that angle AEB will be acute, because it will be subtended by a smaller arc than a semicircle. Form [angle] OEB equal to it, and draw line OE as far as you like. It is evident that [the form of] every point on that line is reflected to A from point E.

2.394◉ Now, if a perpendicular is dropped from point B to line OE, that perpendicular will be equal to, longer than, or shorter than BA. If it is equal, then every line, other than the perpendicular, that is drawn from point B to line OE will be longer than line BA, and so, with one exception [i.e., point O], every point on line OE will lie a different distance from the center than point
A.

2.395◉ If the perpendicular is longer [than AB], then all the points on that line [i.e., OE] will lie farther from the center than point A [and will therefore lie different distances from B than A]. But if the perpendicular is shorter, two lines equal to BA can be drawn from [points on] opposite sides of the perpendicular, whereas all the remaining lines will be either shorter or longer
[than BA and will therefore lie at different distances from B]. It is therefore evident that [the forms of] points that do not lie the same distance from the center as [point] A are reflected to [point] A from point E, which is what was proposed.

2.396◉ From these considerations it is clear that, if [center of sight] A is taken outside the circle—let it be at H [figure 5.2.41a, p. 262]—and if diameter HDBG is drawn along with tangents HT and HQ, then reflection of points that do not lie the same distance from the center as H can take place to H from any point on arc TG other than from T or G. And the proof will be the same [as the
previous one].⁑

2.397◉ [PROPOSITION 42] On the basis of these determinations, it will be established that, if reflection occurs to [point] A from point E or from another point not lying the same distance as point A from the center, the diameter in which the point [whose form is] reflected lies forms two angles with diameter ABG, one of them opposite the reflected angle, the other adjacent to it, and that
adjacent angle will sometimes be greater than the reflected angle and sometimes smaller.⁑

2.399◉ Let it be perpendicular. Accordingly, EA will be parallel to FB, and the two angles FBA and FEA will sum up to two right angles. Now, when line BO is drawn, the two angles OBA and OEA will sum up to less than two right angles. Hence, angle OBG > angle OEA, which is the reflected angle. And since triangle EBF = triangle EBA, BF = BA, and so OB > BA.

2.400◉ Moreover, when line BN is drawn, the two angles NBA and NEA will sum up to more than two right angles. Thus, angle NBG < angle NEA, and NB > BA, and so [the forms of points] N and O will be reflected to [point] A from point E. Also, they lie at different distances than point A from the center, and diameter OB forms an angle with ABG that is greater on the side of G than the
reflected angle [OEA], while diameter NB is longer [than diameter AB], so [we have demonstrated] what was proposed.

2.401◉ If, however, BA is not perpendicular to EA, then draw the perpendicular BK, which either lies above AB [figure 5.2.42a, p. 262], or lies below it [figure 5.2.42b, p. 262]. The proof will be the same [in both cases].

2.402◉ Let BF be perpendicular to EO, draw FT equal to AK, and draw TB. It is evident that in triangle KEB right angle EKB = [right angle] EFB, and angle KEB = angle FEB [by construction]. It follows that the third [angle, i.e., EBK] = the third [angle, i.e., EBF], and since side EB is common to both triangles [EBF and EBK], the triangles will be equal, and FB = KB. But AK = FT [by
construction. And since triangle BAK = triangle BTF] AB = BT, and angle ABK = angle FBT.

2.403◉ With common angle [KBT] added to [both in order to yield angle] FBA, [angle] KBF = [angle] TBA.⁑ But [angles] KBF and FEA
sum up to two right angles,⁑ so [angles] TBA and TEA sum up to two right angles, and so
[angle] TBG = angle TEA, which is the reflected angle.

2.404◉ Therefore, if a line is drawn from point B to [fall on] line ET beyond T, it will form an angle with BG on the side of G that is smaller than the reflected angle. And that line will be longer than AB, because TB = AB.

2.405◉ Furthermore, any line drawn from point B to [fall on] line ET in front of T [i.e., between T and E] will form an angle with BG that is greater on the side of G than the reflected angle, and it will be unequal to AB, so [we have demonstrated] what was proposed.⁑

2.406◉ [PROPOSITION 43] Now, let B be the center of sight and G the center of the sphere. Draw diameter ZBGD [figure 5.2.43, p. 263], and produce the plane that contains the diameter and cuts the sphere along [great] circle ZEH. I say that, if point A is reflected to point B from some point on the circle, and if the distance from point A to the center is not the same as that of point B
from the center, diameter AG will form an angle [AGD] with diameter GD on the side of D that cannot possibly be equal to the reflected angle.

2.407◉ Let it be equal, let T be the point of reflection, and let AG ≠ BG. Draw lines TA, TG, and TB, and construct a circle passing through the three points A, G, and B, and this circle will necessarily pass through point T. For if the circle lies beyond it, and if lines are drawn from points A and B to the corresponding point on that outlying circle, they will form an angle smaller than
angle ATB. And it will be proven that it is equal.

2.408◉ For [angle AGD] along with angle AGB will sum up to two right angles, and angle ATB along with angle AGB sums up to two right angles, since angle ATB = angle AGD, by construction [and within the circle angle ATB + angle AGB = two right angles by Euclid, III.22], and so it is impossible [for T to lie outside the circle]. Likewise, if the circle were to pass below T, the same disproof
would hold [insofar as the new angle ATB would be greater than ATB within the circle].

2.409◉ It follows from this that it must pass through point T, but since angle ATG = angle BTG [by supposition], arc AG = arc BG, and so [chord] AG = [chord] BG. Yet they have been assumed to be unequal, and so [we have demonstrated] what was proposed [i.e., that angle ATB cannot equal angle AGD when AG ≠ BG].⁑

2.410◉ [PROPOSITION 44] Furthermore, if the two points A and B [figure 5.2.44, p. 264] are taken on the two diameters EGH and ZGD so that BG > AG, I say that, if [the form of] point A is reflected to [point] B from two points on arc EZ, the angles of reflection will not both be smaller than angle AGD.

2.411◉ For, on arc EZ select two points, T and Q, from which [the form of point] A is reflected to [point] B, and draw lines BT, GT, AT, BQ, GQ, and AQ. If angle ATB < angle AGD, I say that angle AQB will not be smaller than angle AGD.

2.412◉ Let it in fact be smaller, draw line GN bisecting the angle formed by the diameters [i.e., angle BGA], and draw line AB, which GN intersects at point F. It is evident that BG:GA = BF:FA [by Euclid, VI.3]. But BG > GA; [so] BF > FA.

2.413◉ Bisect AB at point K, and construct a circle passing through the three points A, B, and T, a circle that will not pass through G, because [if it did] angles AGB and BTA [in quadrilateral BGTA] would be equal to two right angles [by Euclid, III.22], and it is obvious that they are less [than two right angles], since angle BTA < angle AGD [by supposition, and angle AGD + adjacent
angle AGB = two right angles]. It will therefore pass above G.

2.414◉ By the same token, it will not pass through Q. For if the point, i.e., M, is taken on the circle where line GQ intersects it, arc AM would be equal to arc MB, since [by Euclid, III.26] they would subtend equal angles [MQA and MQB] at Q, which is impossible, because, if point O, where line GT intersects this circle, is taken, arc AO = arc OB, since they subtend equal angles at T [and
so GM, which supposedly bisects circle BMA, would cut a smaller arc on the side of A than on the side of B]. It follows that this circle passes above Q, for if it passed below, the same disproof would apply.

2.415◉ Now, draw a line from point O to point K [which is the midpoint of BA, by construction], and since it bisects chord AB and likewise bisects arc AB, this line will be perpendicular to AB. But [by Euclid, I.18] angle BAG > angle ABG, since BG > GA. Also, [exterior] angle BFG [of triangle GFA] equals the sum of the two [interior] angles FAG and FGA [by Euclid, I.32], while
[exterior] angle AFG [of triangle BFG] equals the sum of the two [interior] angles FBG and FGB [by Euclid, I.32].

2.416◉ But [angle] AGF = [angle] FGB [by construction], and [angle] FAG > [angle] FBG [because it is subtended by BG, which is longer than AG]. Thus [for the same reason], angle BFG > angle AFG. Accordingly, [angle] AFG is less than a right angle, so [vertical angle] NFB is less than a right angle. But OK forms a right angle on FB. Therefore, when it is extended, it will intersect GN
above BF, never below it.

2.417◉ Now, if a circle is produced to pass through the three points A, Q, and B [figure 5.2.44a, p. 265], it will pass above G, and GQ will bisect its arc A[Q]B [since angles AQG and BQG are presumed equal]. But K bisects chord AB [by construction]. Hence, KO will intersect GN below BF and above point G. Therefore, it will intersect GN sooner [than it should, i.e.,] below FB, and it has
already been disproven [that it can intersect below FB].

2.418◉ It therefore follows that angle AQB may not be smaller than angle AGD, or else [the form of point] A is not reflected to B from point Q. The same disproof will hold if any [other] point is taken on arc EN.⁑

2.419◉ Now, if [some] point C is taken on arc NZ [figure 5.2.44c, p. 267], and if reflection of [the form of] point A to B occurs from point C so that the reflected angle at C is less than angle AGD, just as the reflected angle at T is less than the same angle, [the supposition that the reflected angle at C can be less than angle AGD] is disproved in the following way.

2.420◉ Draw AC, BC, and GC. It follows necessarily that GC intersects KO on arc A[O]B, and line GC will bisect that arc on circle ABT, as will line KO. Accordingly, let point L be where lines GC and KO intersect. When line TC is drawn, then, since the two lines GC and GT are equal [being radii of circle ZHDE], the two angles GCT and GTC will be equal, and both of them will be acute.

2.421◉ Hence, if the line [TX] perpendicular to GT at point T is drawn, it will be tangent to the circle on the mirror, and when it is extended it will fall upon the endpoint of the smaller circle’s diameter [by Euclid, III.31], since the angle it forms with TG is subtended [by an arc equal to] a semicircle on the smaller [circle]. Since TO falls upon KO, and since KO, when extended,
passes through the center of the smaller circle [by virtue of its bisecting line AB along the orthogonal], that perpendicular [TX] will necessarily fall upon the endpoint of KO, when it is extended [by Euclid, III.31], and TC lies below that perpendicular when it is taken with respect to N.

2.422◉ Therefore, no matter what line is drawn [from G] to line TC so as to intersect diameter OK of that circle, it will fall to a point on line TC but below that perpendicular [TX]. Therefore, since GC falls to C and intersects OK, C will lie below the perpendicular and beneath the arc [on circle AOBT passing] through that perpendicular.

2.423◉ Therefore, if a circle is produced to pass through the three points A, B, and C, it will pass through C, and it will intersect circle ABT at the two points A and B. And when it continues past point B, it will pass on to point C, and although it lies below that circle [i.e., ABT], it will necessarily intersect it at a third point, which is impossible.⁑

2.424◉ It follows, then, that [the form of point] A may not be reflected to B from two points on the arc extending between the diameters on which they lie, i.e., arc EZ, in such a way that both angles of reflection are less than angle AGD, which is what was proposed.

2.425◉ [PROPOSITION 45] I say, moreover, that two points lying different distances from the center can be reflected from two points on the arc facing them, that is, the arc lying between the diameters in which those points lie.

2.426◉ For instance, taking two diameters, i.e., BD and GD [figure 5.2.45, p. 268] in the [great] circle of the sphere, let the angle formed by them be bisected by diameter ED, and take point M on BD beyond the point where the perpendicular dropped from point E to BD will fall. Then take ND equal to MD, and construct a circle passing through the three points D, N, and M. That circle will
necessarily pass beyond E, for if [it passed] through E, it would create a quadrilateral at the four points D, N, E, and M, and the two opposite angles of that quadrilateral are equal to two right angles, which would not be so [in this case], since line EM lies beyond the perpendicular, and angle EMD is acute.

2.427◉ Likewise, its counterpart at N is acute, because EN [also lies] beyond the perpendicular [so EMD and END sum up to less than two right angles]. The disproof will be similar if the circle is [assumed] to pass below E. Therefore, it will pass beyond, and it will intersect the [great] circle of the sphere at two points, e.g., T and L.

2.428◉ Draw lines MT, DT, NT, ML, DL, and NL, and draw line MN to intersect [line] TD at point F and line ED at point P. It is evident that, since MD = ND [by construction], since PD is common, and since the [subtended] angle [MDP] = the [subtended] angle [NDP, by construction], triangle [MDP] = triangle [NDP], and angle FPD will be a right angle. Hence, angle PFD is acute.

2.429◉ From point F draw KF perpendicular to TD. It is clear that any point on line NL will lie below point K [since FK intersects ML between M and L]. Taking the position of [some such] point [on NL] below [K] with respect to T, let that point be Z, and draw line TZ until it reaches point C on the circle. Arc NC is either shorter than arc TL or not.

2.430◉ If it is not shorter, then take an arc on it [i.e., on arc NC] that is shorter, and draw a line from point T to the endpoint of that arc, and it will be the very line [TC that we wanted in the first place].

2.431◉ So let NC < TL. It is obvious that angle TNL > angle CTN, since it is subtended by a longer arc [i.e., TL as opposed to CN]. Cut an [angle] equal [to CTN] from it, let that angle be INZ, and at point T form angle OTM equal to angle CTN. Since angle TML > angle MTO,⁑ line TO will intersect line LM. Let it intersect at point O.

2.433◉ But TZ > TO, which becomes clear as follows. Let R be the point where TZ intersects KF. Angle TFR is a right angle [by construction], so angle FTR is acute. Hence, angle OTF, which equals it, is acute.⁑ Moreover, KF is perpendicular to TD [by construction], so when it is extended it will intersect TO, and the line [TX] drawn from point T to that point of
intersection, a line that includes TO as a segment, will be equal to line TR [because triangles TRF and TXF are equal, by Euclid, I.26]. And thus TO < TZ, so NT:TO > NT:TZ.

2.437◉ Accordingly, [the form of point] O is reflected to Z from the two points T and L [given that OTD = ZTD, by previous conclusions, and ZLD = OLD, since they are subtended by equal arcs], and O and Z lie unequal distances from the center, and they lie on different diameters.

2.438◉ That they do not lie on the same diameter is evident from the fact that angle SDN = angle ODM. Thus, when the common angle SDM is added [to each] angle, [angle] NDM = angle SDO. But angle NDM < two right angles, so angle ZDO < two right angles [which means that Z, D, and O cannot lie on a single straight line]. Therefore, O and Z do not lie on the same diameter but on
different ones.

2.439◉ [PROPOSITION 46] Furthermore, given two points O and K [figure 5.2.46, p. 269] that lie different distances from the center [of the mirror], one will be reflected to the other from two points on the arc facing the radii in which those point lie, but they will not reflect from any point on that arc other than those two.⁑

2.440◉ For example, let D be the center [of a great circle on the mirror], let K lie farther from D than O does from D, let GD and OD be diameters, and let T be one point of reflection. It is clear from earlier discussions [in propositions 43 and 44 above] that the two reflected angles will not [both] be smaller than, nor equal to, angle ODA. Hence, one of them will be greater. Let the
reflected angle at point T be greater [than angle ODA], and draw lines OT, DT, and KT.

2.441◉ Then, from that reflected angle [OTK] cut off [angle] OTF equal to angle ODA, and bisect angle FTK with line TE. From point K draw [line KZ] parallel to TF, a line that will intersect TE. Let it intersect at point Z, draw line OK, bisect angle ODK with line DU so that it intersects line OK at point C, and let KD > OD. Therefore, since KD:DO = KC:CO [by Euclid, VI.3], KC > CO.
Then, let line DT intersect line OK at point N. I say that C lies between N and K, not N and O, which will be shown as follows.

2.443◉ If a circle is constructed through the three points O, T, and K, it will pass below D, because, since angle OTK > angle ODA [by construction], the two angles OTK + ODK > two right angles, and line ND will bisect arc OK of that circle below D.⁑

2.444◉ If a line is drawn from the point of bisection [s] to the midpoint [x] of line OK, which is the chord on that arc, that line will be perpendicular to OK [by Euclid, III.3], and it will fall between C and K, since CK > CO [by previous conclusions]. Moreover, the angle at [point] N beyond that perpendicular on the side of C [i.e., DNC] will be acute, and the angle at C on the side
of O [i.e., OCD] is acute [since angle KCD is obtuse, by previous conclusions]. Accordingly, if C were to fall between N and O, it would be impossible for that perpendicular to fall between N and C, because it would intersect DC and would form a triangle with one angle right and the other obtuse.⁑

2.445◉ Hence, it [i.e., the perpendicular] will fall between N and K, and the angle at N on the side of the perpendicular will be acute, so this [same angle] will be obtuse on the side of C [if C lies between N and O], and so there will be a triangle with two obtuse angles.⁑

2.448◉ Thus, if a perpendicular is dropped from point K to TZ, it will fall between E and Z. For if it were to fall above E, then, since angle TEK is obtuse, it would follow that the triangle [formed by KD and the perpendicular intersecting TE above E] would have two [of its three] angles [consisting of] a right angle [formed by the line from K that intersects TE above E] and an obtuse
angle [KET]. Let KQ be the perpendicular, then. I say that KT:TF = KD:DO.

2.450◉ If they are parallel [figure 5.2.46a, p. 270], then, since they fall between parallels, they will be equal. If, however, they intersect [figure 5.2.46b, p. 270], then they form a triangle [two of] whose sides [OP and TP] are equal, because they subtend equal angles [since angle DOT = alternate angle ODA, and angle OTF = angle ODA, by construction], and because FD intersects those
sides parallel to the base [TO]. Therefore, the ratio of one side to DO will be the same as the ratio of the other side to TF, and so TF = DO.

2.451◉ {I claim that this is so if they [i.e., DO and TF] intersect below KD. And if they intersect below TO, the same proof will apply, because they will form a triangle one of whose sides is TO and the other two of which are equal, and the ratio of one [of those equal sides] to DO will be the same as the ratio of the other to TF.}⁑ Furthermore, angle TDK = [alternate] angle DTO, because DT lies between parallels. Thus, it [i.e., angle TDK] = angle DTK [which = angle of reflection DTO], so DK and TK are equal. Hence, TK:TF = KD:DO.

2.452◉ If, on the other hand, TO intersects KD, let it intersect at point P on the side of A [figure 5.2.46c, p. 270]. We know that KT:TF is compounded of KT:TP and TP:TF [i.e., KT:TF = (KT:TP):(TP:TF)]. But KT:TP = KD:DP [by Euclid, VI.3], because DT bisects angle KTO. Moreover [given that triangles TPF and DPO are similar], TP:TF = DP:DO, because angle ODP [i.e., ODA] = angle PTF [by
construction], and the angle at P is common. Part [ODP] of the [larger] triangle [PTF] is [therefore] similar to the whole. Hence, KT:TF is compounded of KD:DP [which = KT:TP] and DP:DO [which = TP:TF]. But KD:DO is compounded of the same [ratios, i.e., KD:DP and DP:DO], so KT:TF = KD:DO.

2.453◉ If, however, TO intersects KD on the side of G [figure 5.2.46d, p. 270], let L be the [point of] intersection, and from point D draw [line] DR parallel to line KT so as to intersect TO at point R. Accordingly, angle KTD = [alternate] angle TDR, but it is [also] equal to angle DTO [by supposition], so DR = TR. However, since triangle LTK is similar to triangle LRD [because TK and RD
are parallel], DR:RL = KT:TL, and so RT [which = DR]:RL = KT:TL. But RT:RL = DK:DL. Thus, KT:TL = KD:DL.

2.456◉ Since KQ is perpendicular to EZ, though, all of its angles [of intersection with TZ] will be right angles. But angle ETD is acute, because it is half of angle [FTO, by earlier conclusions]. Therefore, KQ will intersect TD. Let H be the intersection, draw line EH, and from point E draw EC’ parallel to KH, and extend it to DH [which it intersects at point C’].

2.457◉ Adjust the figure according to the interrelationship of lines [figure 5.2.46e, p. 271], and construct a circle that passes through the three points C’, T, and E. Extend KD until it reaches the circle at point M, and draw MT. Angle TME = angle TC’E, since they are subtended by the same arc, and angle TC’E = angle C’HK [since EC’ and KH are parallel, by construction]. [Therefore,
angle] TME = angle C’HK.

2.458◉ Cut from angle TME angle F’MD equal to angle DHE, and let I be the point where F’M intersects TC’. It is evident that triangle IMD is similar to triangle EHD [because all their corresponding angles are equal], so HD:DM = EH:IM.

2.460◉ But KD:DT is known, since it remains one and the same throughout, no matter where point of reflection T might lie on arc EG, because TD remains unchanged, and so does KD. Line EH also remains one [and the same] no matter the reflection, so it does not change its length, and so line IM will always be one [and the same], and so point F’ is known and determinate.⁑

2.461◉ Therefore, if reflection could occur from three points on arc BG, three lines equivalent [to FM] could be extended from point F’ to circle TC’E, because KD:DT would be as EH is to any one [segment IM] of them. But it is clear from earlier discussion [in proposition 20, lemma 2 above] that only two equal lines can be [so] drawn, so reflection will occur from only two points, which is
what was proposed.⁑

2.463◉ For example, take line ZT’, and cut it at point E so that ZE:ET’ = KD:DO. Since KD > DO [by construction from the previous proposition], ZE > ET’. Bisect ZT’ at point Q, from point Q draw a line [K’QH] perpendicular to ZT’, and form angle ET’D’ equal to half of angle ODA. It will be acute. Accordingly, T’D’ will intersect the perpendicular [KQ].

2.464◉ Let the intersection be at point H, and [by proposition 24, lemma 6] draw line D’EK’ so that K’D’:D’T’ = KD:[DT], the radius of the sphere. Then, form angle KDT in the mirror equal to angle K’D’T’ that we have [in the construction based on T’Z]. I say that T is a point of reflection, and if you repeat the previous proof, you will see this clearly.⁑

2.465◉ [PROPOSITION 48] Moreover, if two points are taken on different diameters, if they lie at different distances from the center, and if they lie outside the circle and are reflected from some point on the [concave] arc facing the diameters, they will not be reflected from any other [point] on the same arc.

2.466◉ For example, let A and B [figure 5.2.48, p. 276] be points lying outside the circle on different diameters, G the center [of the great circle on the mirror], and T the point of reflection, and draw BT, AT, and GT. BT will intersect the [convex] arc on the circle. Let Q be the point of intersection. So too, AT will intersect the [convex] arc on the circle. Let M be the point of
intersection.

2.467◉ Since angle BTG = angle ATG [by construction], they are subtended by equal arcs on the circle, which will be evident if diameter TG is drawn. Accordingly, arc QT = arc MT [and so, therefore, do chords QT and MT]. Thus, if [the form of point] B is reflected to [point] A from a point other [than T], let it be H, and draw lines BH, AH, and GH. Let BH intersect the circle at point L, AH
at point N.

2.468◉ According to the previous reasoning, HL = NH [because they subtend supposedly equal arcs]. But we have just [established] that QT = TM, which is impossible [if HL = NH]. It follows that [the form of point] B may not reflect to [point] A from point H or from any point other than T on the arc facing [either] diameter.

2.470◉ Moreover, if the line extending from one of the two points [A or A’ in figure 5.2.48a, p. 276] to the other [B] is tangent to the circle or lies entirely outside it, then, if some point [T] is taken on the [convex] arc facing the diameters, one of the lines [i.e., either AT or A’T] extended from [each of] the two points to that point will lie entirely outside the circle [because it
will strike its outer rather than its inner surface]. And so neither of the points [A and B, or A’ and B] will be reflected to the other from any point on that arc [CE], and [it will be reflected] from only one point on the opposite arc [KD] of the mirror, and so [it will be reflected] from only one point on the [entire] mirror.

2.471◉ [PROPOSITION 49] On the other hand, if the line extending from one point to the other [i.e., from object-point to center of sight] cuts the [great] circle [on the mirror], construct the circle [passing] through the center of the mirror and those two points. That [second] circle will lie entirely within the circle [of the mirror], or it will touch it [at one point], or it will
intersect it.

2.472◉ Let it lie entirely within [figure 5.2.49, p. 277], and draw two lines [AT and BT] from the two points to some point [T] on the facing arc [of the mirror]. The angle they will form [ATB] will be smaller than the angle one diameter forms with the other on the adjacent side of the [mirror’s] center [i.e., angle DGB]; and no matter what angle is formed in this way on the facing arc, it
will be smaller than the latter angle.

2.473◉ For the angle [ACB] formed within the inner circle by the lines drawn from the points to the arc on it lying between [the two points] will be equal to that latter angle [DGB], because, combined with the angle [AGB] formed by the diameters above the center, it sums up to two right angles [by Euclid, III.22]. But the angle [ACB] within arc [ACB] in the smaller circle is greater than
the angle [ATB] within arc [CTH of the circle] on the mirror.

2.474◉ Therefore, in the arc of the circle [on the mirror], reflection will occur from only one point, since it has already been claimed [in proposition 44 above] that it is not possible for reflection to occur from two points such that both [reflected] angles are smaller than the angle formed by the diameters on the adjacent side of the center.

2.475◉ If, however, that [second] circle touches the circle of the mirror at one point [T in figure 5.2.49a, p. 277], the angle formed by the lines drawn from those points [A and B] to the point of contact [T] will be equal to the angle [DGB] formed by the diameters on the adjacent side of the center [because AGB + DGB = two right angles, by Euclid, III.22, and so do AGB + ATB], so no
reflection will occur from that point of contact [according to proposition 43 above]. Moreover, the angle formed at any other point [e.g., T’] on the arc [CH] of the larger circle will be smaller than that one, so, according to previous claims [in proposition 44 above], reflection will not occur from two points on that arc.

2.476◉ If, on the other hand, the inner circle cuts the circle of the mirror, the two points [A and B] will lie outside the circle [of the mirror], or [they will both lie] inside it, or one [will lie] inside and the other outside, or one [will lie] on the circle and the other outside or inside it.

2.477◉ If they [both] lie outside [as represented by A and B in figure 5.2.49b, p. 277], or if one lies on the circle and the other outside [as represented by A and B’ in the same figure], then the [second] intersecting circle will not cut the arc of the mirror’s circle between the diameters, and so, no matter what angle is formed on that arc [e.g., ATB or ATB’], it will be larger than the
angle [DGB] formed by the diameters on the adjacent side of the center. And it has already been demonstrated in the preceding figure [i.e., proposition 48] that [the forms of] these points can be reflected from only one point on the [concave] arc [KD] lying between [the diameters].⁑

2.478◉ But if the two points lie inside [the mirror’s circle, as represented in figure 5.2.49c, p. 277], the inner circle will cut the arc lying between [the diameters] at two points [E and F], and there will be two arcs [CE and FH] left over on opposite sides [of arc CH facing the diameters].

2.479◉ If one of the points lies inside the circle and the other on the circle or outside it [as represented in figure 5.2.49d, p. 278], the [second] circle will cut the arc [CB’] lying between [the diameters] at a single point [E], and only one arc-[segment, CE] will be left over.

2.480◉ If it intersects [the arc between the diameters] at two points, all the angles [e.g., ATB in figure 5.2.49c, p. 277] formed upon the arc lying between the two points of intersection [E and F] will be greater than the angle [DGB] formed by the diameters on the adjacent side of the center, and from this arc [EF] reflection may occur from only one point, or it may occur from two [by
proposition 46 above].

2.481◉ And from the two arcs [CE and FH] that are left over from the entire arc on opposite sides of [points E and F of intersection], all the angles [e.g., AT’B] will be smaller than the angle [DGB] formed by the diameters [on the adjacent side of the center], and reflection will occur from only one point on them [by proposition 44 above].

2.482◉ And [so] in this case [when two arc-segments are left over on opposite sides of intersection-points E and F], reflection can occur from two points on the arc lying between the diameters, or from three points [in the whole arc CH, i.e., two from EF and one from CE or FH].

2.483◉ Moreover, it is clear [from proposition 38 above] that reflection will occur from only one point on the opposite arc [DK], so in this case it may occur from three [points], or it may occur from four.⁑

2.484◉ But if it [i.e., the second circle] cuts the arc lying between the diameters at only one point on the larger circle, all the angles formed on the segment of that arc [EB’ in figure 5.2.49d, p. 278] included within the smaller circle will be greater than the angle [DGB’] formed by the diameters [on the adjacent side of the center], and [so] reflection can occur from two points, or
from one point, on that part [CB’ of the arc facing the diameters].

2.485◉ All the angles in the other side [CE] of the arc lying between [C and B’] will be smaller than the angle formed by the diameters [on the adjacent side of the center], and reflection will occur from only one point on that segment, and so, given that reflection always occurs in this case from one point on the opposite arc [DK of the mirror], reflection may occur from three [points],
or it may occur from four, but never from more.

2.486◉ It is therefore clear that points lying different distances from the center may at times be reflected from only one point, at times from two, at times from three, or at times from four, but never from more [than four]. Moreover, if the [two] points lie the same distance [from the center], reflection can occur from one point only, or from two, or from four, [but] never from three
[alone].⁑

2.487◉ When reflection occurs from one point, one image appears; when [it occurs] from two, two [images appear]; when [it occurs] from three, three [images appear]; when [it occurs] from four, four [images appear]. But if the visible point and the center of sight lie on the same diameter, reflection will occur from an entire circle [within the sphere of the mirror], and the image-location
will be [at] the center of sight [by proposition 36 above]. If the center of sight lies at the center of the mirror, though, it sees nothing [other than itself]. On the other hand, if the visible point lies at the center of the mirror, it[s image] will not be seen, because its form will reach the mirror along the normal and can be reflected only along the normal.

2.488◉ Yet when the center of sight and the visible point lie outside the center on different lines, those lines, when extended to the center, will cut two arcs on different sides of the circle within the sphere. Reflection will occur from only one point on one [of those arcs], but [it may occur] from three [points] on the other. But if the center of the sphere lies on one side, while the
center of sight and the visible point lie on the other, the arc that the diameters cut [on one side or the other in the circle] will be blocked by the [viewer’s] head, so reflection will then occur from only three points [at most]. And if in that case the eye is directed to the arc where only a single reflection occurs, the other arc [from which] three [reflections occur] will be blocked, and only one image will appear.

2.490◉ When something is perceived in this [sort of] mirror with both eyes, if the line of reflection is parallel to the normal, the image-location will lie at the point of reflection [by proposition 32 above], and since the points of reflection are separated from one another with respect to the two eyes, two images of the same point will appear to the two eyes [and will be melded at the
point of reflection]. On the other hand, if the line of reflection is not parallel to the normal, and if the visible point lies the same distance from one eye as from the other, or if the difference [in distance] is slight, the image-location will be the same for both eyes, or it will be different [for each eye] but only slightly divergent. Therefore, either a single image, or virtually a single image, will appear, as was demonstrated in the case of convex
spherical mirrors [in paragraph 2.221 above].

2.491◉ In the case of concave cylindrical mirrors, the common section [of the plane of reflection and the mirror] is sometimes a straight line. When the plane of reflection intersects the axis, the common section is sometimes a circle—[i.e.,] when that plane is parallel to the bases [of the cylinder]—[and] the common section is sometimes a cylindric section [i.e., an ellipse]. When [the
common section] is a straight line, image-location and the analysis of reflection will be the same as in plane mirrors. When it is a circle, the analyis will be the same as in concave spherical [mirrors]. However, when the [common] section is cylindric [i.e., elliptical], the image-location will lie behind the mirror, or beyond the center of sight, or at the center of sight, or between the mirror and the center of sight, or on the mirror itself, which will be
demonstrated in the following way.

2.492◉ [PROPOSITION 50] Let ABG [figure 5.2.50, p. 280] be the [elliptical common] section. Draw normal DG within this section, and, according to previous discussion [in book 4], it is clear that this normal is a diameter of the circle [parallel to the cylinder’s base and coincident with the section], and it must be unique, because from no other point on the section can a normal be drawn
to the plane tangent [to both the circle and the section].140 Select another point [on the section], let it be B, and from it draw a line within the section that is normal to the line tangent to the section at point B, and, as claimed earlier [in proposition 26], this line will necessarily intersect normal [GD]. Let it intersect at point D, and let [point] B be chosen near enough to point G that angle BDG is acute.

2.493◉ Then, from point G, draw line GH parallel to BD within the section, and it should lie within the cylindric section, because angle HGD will be acute, since it is equal to [alternate angle] GDB. From point G draw a line between D and H, and it will necessarily intersect BD. Let it intersect at point N, and between N and G select some point O. Beyond point N [on line GN] select point
T. Furthermore, from point G draw another line GZ above GH, [but] still within the section, and it will necessarily intersect BD on the other side [of G]. Let E be the [point of] intersection. Draw line GQ so that angle QGD = angle ZGD, and form angle LGD equal to angle HGD and angle MGD equal to angle NGD.

2.494◉ It is evident that, if the center of sight lies at point Z, [the form of] point Q will be reflected to it from point G [since QGD = ZGD by construction], and point E [behind the mirror on normal QD will be the location] of its image. If the center of sight lies at point H, [the form of] point L will be reflected to it from point G [since LGD = HGD by construction], and its
image-location will be [point] G [on the mirror’s surface, because normal LD is parallel to line of reflection GH]. If the center of sight is at point O, [the form of] point M will be reflected to it [from point G, since MGD = NGD by construction], and its image-location will be [point] N [behind the eye on normal MD]. If [the center of sight] lies at N, the image-location [for the form] of point M will be at the center of sight [itself], i.e., at N [where normal
MD intersects line of reflection GN]. And if [the center of sight] is at T, the image-location [for point M] will lie between the eye and the mirror, because it lies at N, and so [we have demonstrated] what was set out [to be proven].

2.495◉ These conclusions must be understood [to apply] when the visible point does not lie on the [same] normal as the center of sight, for in that case, since an infinite number of planes can all be imagined to lie on that normal such that each one is orthogonal to the plane tangent to the mirror, and since all of them lie on the normal, any one of those planes will form a rectilinear
common section [with that tangent plane]; and reflection will only occur along the same normal, the center of sight [will constitute] the image-location, and no point will be seen unless it lies on the surface of the eye.

2.496◉ However, one of those planes forms a circular common section [with the mirror], and in that case, when the center of the mirror [D in figure 5.2.50a, p. 281] lies between the visible points [e.g., C] and the eye [E], each of those points can be reflected to the eye from two points [e.g., H and L] on the circle, since lines may be drawn from each of them to form an angle with the
plane tangent [to the point of reflection such] that the diameter [HDL] drawn [from that point of reflection] to the center [of the circle] bisects [that angle]. I say this, of course, about points that lie on that normal, and their image-locations lie at the center of sight.⁑ The other points on that normal [i.e., between D
and E in figure 5.2.50a] will not be reflected to the eye except for the point that lies on the surface of the eye, and that one [is reflected] along the normal.

2.497◉ On the other hand, when the common section is a cylindric section, the points on the normal cannot be reflected from any [other] points on the section, because the form reaching along the normal must be reflected along the normal, and in [such] a section the normal is unique [as shown in book 4], so reflection will occur only along this normal, and only the point on the surface of
the eye [will be so reflected], and the image-location will be at the center of sight.

2.498◉ If, however, the center of sight lies at the center of the circle, the portion of the eye that the normals extending from the center of sight to the circle [on the mirror] cut off will be reflected from the corresponding portion of the circle [on the mirror] that the normals cut. Since any line extending from the center of sight to the circle is normal, reflection will occur along
the normal, and the image-location will be [at] the center of sight, which is the center of the circle.

2.499◉ Now, at point A [figure 5.2.50, p. 280] form acute angle FAG of some kind. It is clear that FA will intersect GZ. Let the intersection be at point Z, and form angle CAG equal to angle FAG. AC will intersect GQ. Let the intersection be at point C. It is evident that [the form of point] C is reflected to [point] Z from point G, and it is also reflected to [point] Z from point A, but
not from any other point on the section, because it cannot reflect except from the endpoint of the normal, and there is only one such normal in the section, namely, GA.

2.500◉ [PROPOSITION 51] Moreover, if two points are taken on the axis of the cylinder, [the form of] one will be reflected to the other from one full circle in the cylinder, and the image-locations will lie on a given circle outside the cylinder.

2.501◉ For example, Let EZ [figure 5.2.51, p. 282] be the axis, T and H two points selected on the axis, and AG and BD the bases of the cylinder. Bisect TH at point Q, and construct a circle with Q as its center, i.e., LM, that will be parallel to the [cylinder’s] bases, LM being its diameter, and BLA and DMG being sides of the cylinder. Also, construct circle KC with H as its center and
CK its diameter, and draw lines TL, TM, HL, and HM.

2.502◉ It is evident that each of the four angles at Q is a right angle, that TQ = QH, and that QL = QM. Those triangles [i.e., TLM and MLH] will be similar, so angles TLQ and QLH will be equal; likewise, angles TMQ and QMH will be equal. Therefore, if H is the center of sight, [the form of] point T will be reflected to point H from point L, and likewise from point M. Accordingly, if
triangle TLH is rotated while axis TH remains stationary, point L will describe a circle, the two angles TLQ and QLH will remain constant throughout, and throughout this motion [the form of] T will be reflected to H.

2.503◉ Now, draw line CHK until it intersects line TL, and let F be the [point of] intersection. It is evident that F will be the image-location, and as triangle TLH revolves, triangle TFH will revolve, and during this motion point F will describe a circle outside the cylinder. That entire circle will be the location for [all] images, and this is what was proposed.⁑ The same method of proof will apply for any two points [chosen] on the axis.

2.504◉ [PROPOSITION 52] Furthermore, some points that are selected outside the normal [extending] from the eye have one image, some [have] two, some [have] three, and some [have] four, but none [has] more [than four].

2.505◉ For instance, let A [figure 5.2.52, p. 283] be a visible point outside the normal [extending] from the center of sight, and construct a plane passing through A and parallel to the bases of the mirror. It will, of course, form a circle on the cylinder. Let H be the center of that circle, and choose another point B within the plane of the circle, and draw diameters AH and BH [B thus
serving as a center of sight within this plane].

2.506◉ From what has been claimed about spherical concave mirrors [in proposition 49 above], it is clear that [the form of point] A may be reflected to [point] B from [at least] one point on the arc [passing through points E, D, and G] that subtends those two diameters, [or] perhaps from two or three, but from no more [than three]; and from the opposite arc [reflection can occur] from only
one point. Accordingly, let [the form of point] A be reflected to [point] B from three points on the subtending arc [that passes through points E, D, and G], let those points be G, D, and E, and draw lines AG, HG, BG, AD, HD, BD, AE, HE, and BE.

2.507◉ From point A in the same plane draw three lines AK, AF, and AN parallel to the three diameters HG, HD, and HE [respectively]. Hence, since AK is parallel to HG, BG will intersect AK. Let it intersect at point K. Likewise, BD will intersect AF. Let the intersection be at point F. So, too, BE [will intersect] AN. Let the intersection be at point N.

2.508◉ Then, from point H erect axis HU, and from point B erect a line perpendicular to the plane of the circle. Let it be BT, and it will be parallel to the axis. Take some point T on it, draw the three lines TK, TF, and TN, and from the three points G, D, and E, erect three lines GM, DL, and EQ [respectively] perpendicular to the plane of the circle [each line thus being a line of
longitude on the cylinder’s surface]. They will be parallel to TB. Hence, EQ will lie in the plane of triangle TBN. So EQ will intersect TN. Let it intersect at point Q. Let DL intersect TF at point L, and [let] GM intersect TK at point M. These three perpendiculars will constitute lines of longitude on the cylinder.

2.509◉ From point Q draw a line parallel to line NA, and it will intersect axis UH, because it will be parallel to EH [which is parallel to NA by construction]. Let the intersection be at point U, and draw line TA, which QU will intersect, because QU is drawn from one side of the triangle [i.e., TNA] parallel to the base [AN, which is parallel to HE by construction]. Let I be the point of
intersection, and draw line QA.

2.513◉ Nor can it be reflected from more points, for let another [such point] be given. If a side [i.e., a line of longitude] is drawn from that point, it will fall on the circle that we have, and, by repeating the proof, it will be demonstrated [according to proposition 49 above] that from the point where the side falls to the circle it is impossible for [the form of point] A to be
reflected to [point] T.

2.514◉ [But the form of point] A can be reflected to [point] B from one point on the opposite arc of the circle. Let Z [in figure 5.2.52, p. 283] be that point, and draw diameter HZ as well as line AC parallel to it. Then draw BZ, and let it intersect AC at point C. Erect perpendicular OZ, which will be a line of longitude and [thus] parallel to TB, and draw TC, which will be intersected
by line OZ. Let the intersection be at point O. It will be proven according to the previous method that [the form of point] A is reflected to [point] T from point O.⁑ And if another point from which reflection can [supposedly] occur is chosen on that side of the cylinder, it will be demonstrated by repeating the proof [according to which the line of longitude is dropped from that point to the circle] that it is impossible for [the form of point A] to be reflected [to point B] from any point on that side of the circle other than Z [as demonstrated in proposition 38 above].

2.515◉ Therefore, if [the form of point] A is reflected to [point] B from one point on a given side of the circle, it is reflected from one [point] on the same side of the cylinder; if [it is reflected] from two [points on the circle, it will be reflected] from two [points on the cylinder]; if [it is reflected] from three [points on the circle, it will be reflected] from three [points on
the cylinder]; but [such reflection] can [occur] from no more [than three points on that side]; whereas on the opposite side [it can occur] from only one point on the circle and [thus] from only one point on the cylinder.

2.516◉ Furthermore, TB is parallel to UH, and there can be no plane chosen, other than plane TBUH, in which T lies with UH. Likewise, there can be no plane other than AUH in which A lies with UH, and it is perpendicular [to the plane of the base-circle]. Hence, T does not lie in the same perpendicular plane with A, nor on the same circle [forming its plane], nor is it on the axis, because
it lies on a line parallel to it. Accordingly, the plane in which [the form of point] A is reflected to [point] T constitutes a cylindric section [because it is oblique].

2.517◉ Now, let TA be extended beyond T and A on both sides to form RP. Since there are four planes of reflection, because [reflection occurs] from four points, and since the two points T and A lie in each of them, RP will be common to the four planes of reflection. Moreover, each of these planes cuts the plane tangent to the mirror at a point on its own common section [with the mirror],
but not on the same common section [as any other plane of reflection].⁑ Line RP is normal to one of the four common sections, but not to two, for if it were perpendicular to the tangent plane [common to two or more common sections], it would thus reach the axis. Hence, the normals [dropped] from point T to these four common sections are distinct [from one another], and there is only one that passes through A.⁑

2.518◉ Moreover, the normal [within any of the planes of reflection] will either be parallel to the line of reflection or will intersect it beyond or inside the mirror. If it is parallel, the point of reflection will be the image location, as has been demonstrated [in proposition 32], and since there are four points of reflection, there will be four images. If it intersects, then, since
there are four normals there will be four intersections and four images.

2.520◉ In concave conical mirrors, the common section of the plane of reflection and the surface of the mirror will be a line of longitude along the mirror, or it will be a conic section. If it is a line of longitude, the image-locations will lie in [i.e., behind] the mirror[‘s reflecting surface].⁑ If it is a conic section, the image-locations will sometimes lie beyond the center of sight, sometimes at the center of sight [itself], sometimes between the center of sight and the mirror, and sometimes behind the mirror, just as was shown in the case of the concave cylindrical mirror.

2.521◉ Furthermore, if a physical spot is taken on the normal extending from the center of sight to the plane tangent to the mirror [and if it lies] between the center of sight and the mirror, its form will not be reflected to the center of sight along the normal, because that spot will block the endpoint of the normal [at the eye], and for that reason it will not be reflected from it.
However, if there is no such [physical] spot on that normal, [the form of] a visible point will be reflected to the eye along this normal, that point, and that point only, being the one on [the surface of the eye] that intersects the normal.

2.522◉ On the other hand, if the center of sight lies on that normal as well as on the axis, it will form a circle, and the line drawn to any point on it from the center of sight will be normal to the plane tangent [to the mirror at that point], so from any point on that circle reflection can occur to the eye along the normal. And the portion of the eye that the two normals cut off to form
the greatest [visible] angle on it will be reflected.

2.525◉ For instance, let H [figure 5.2.53, p. 286] be the center of sight, and T the visible point. Construct a plane that cuts the cone along the length of the axis, and let it be ABGH, with AH the axis and AB and AG edges of the cone. From point T draw TQ perpendicular to line AB, and extend it to QL. Let [QL] = QT. Then, from point H draw a line to point L, and it will intersect line of
longitude AB. Let it intersect at point B, and from point B draw a line parallel to line TQ, which will necessarily reach the axis. Let it reach [the axis] at point D, and draw line TB.

2.526◉ Since TQ is perpendicular to AB, and since TQ = QL, it is clear that triangle BTQ = triangle BQL, and angle QLB = angle QTB. But angle QTB = [alternate] angle TBD, and angle DBH = [alternate] angle QLB. Therefore, angle TBD = angle DBH, and so [the form of point] T is reflected to H from point B, and L is the image-location.

2.527◉ Accordingly, if triangle TLH is rotated [about axis TH], point B will describe a circle on the cone, and from any point on that circle [the form of point] T will be reflected to [point] H. Meanwhile, outside this circle, L will describe a circle that will constitute in its entirety the image-location for point T.

2.528◉ [PROPOSITION 54] Now, in this [sort of] mirror, having selected two points, i.e., Z and E [figure 5.2.54, p. 287] outside the normal [extending] from the center of sight and outside the axis, construct a plane on [point] Z parallel to the base [of the cone]. It will produce a circle in the mirror. E will lie in this circle or in another plane.

2.529◉ Let it lie in the plane of that circle, and draw line EZ. It is evident [from proposition 49, case 3 above] that [the form of point] Z is reflected to E on one side of that circle from one point, or from two, or from three; and on the other side [it is reflected] from one [point].

2.530◉ Take a point on the circle from which [the form of point Z] is reflected [to E], let it be H, and [let] T [be] the center of the circle. Draw lines ZH and EH. Diameter TH will bisect the angle [formed by them], and it will intersect line EZ. Let it intersect at point Q, let A be the vertex of the cone and AH a line of longitude.

2.531◉ From point Q draw line QM falling orthogonally to line AH, and let it reach the axis [AT]. Let it fall at point D on the axis, and draw lines ZM and EM. From point Z in the plane of the circle draw line ZL parallel to line QH. Let EH intersect it. Let the intersection be at point L, and from point H draw HC perpendicular to LZ.

2.534◉ CN is parallel to OL [by construction, so, by Euclid, VI.2] LC:CZ = ON:NZ, so ON = NZ. Furthermore, since OZ is parallel to QM [by construction], plane ZLO will be parallel to plane QMH. Plane EOL intersects these two [planes] along [rectilinear] common sections, i.e., MH and OL, that will be parallel, so HM and CN are parallel [because CN is parallel to OL, by construction]. And
because HC falls between parallels LZ and HQ, and since it is perpendicular to LZ, it will be perpendicular to HQ, so CH will be tangent to the circle.

2.536◉ But EM:MO = EH:HL [by Euclid, VI.2], EH:HL = EH:HZ [by Euclid, V.7, because HL = HZ by previous conclusions], and EH:HZ = EQ:QZ [because HQ bisects angle EHZ]. Thus, EM:MZ = EQ:QZ [since EM:MZ = EM:MO = EH:HL = EH:HZ], so [because the respective sides of triangles EMQ and ZMQ are proportional, making the two triangles similar] angle EMQ = angle QMZ, [and] so [the form of point] Z is
reflected to E from point M. Hence, if [the form of point] Z is reflected to E from point H on the circle, it is reflected to the same point from point M on the cone. And if [it is reflected] from two [points] on the circle, [it is reflected] from two [points] on the cone; if [it is reflected] from three [points on the circle, it is reflected] from three [points on the cone]; and if [it is reflected] from more [points on the circle, it is reflected] from more
[points on the cone].⁑ On the other side of the circle, the proof that [reflection occurs] from one point on the cone just as from one [point] on the circle
will be constructed in the same way.

2.537◉ However, if E does not lie on the circle that passes through Z parallel to the base [of the cone], E will lie above or below it. Let it lie above, since the same proof applies to both cases. Draw line AE [figure 5.2.54a, p. 288] until it touches the plane of that circle, and let H be the point of contact, Q being the center of the circle. It is obvious that [the form of point] H can
be reflected to Z from some point on the circle. Let it be T, and draw diameter QT [normal to the point of reflection]. Line HZ will intersect this diameter at point N. Draw [line] EZ and line of longitude AT.

2.538◉ Since point Z lies on one side of diameter QT, and since [point] E lies on the other, it is evident that line EZ will intersect plane AQT. Let it intersect at point O, and from point O draw [line] OC perpendicular to line AT, and it will necessarily fall upon the axis. Let it fall at point D, and draw lines EC and ZC. I say that [the form of point] E is reflected to Z from point
C.

2.539◉ [Here is] the proof. From point Z draw line ZF parallel to [line] QT, and extend line HT until it intersects it. Let the intersection be at point F. Likewise, from point Z draw [line] ZK parallel to line OC, and extend line EC until it intersects it. Let the intersection be at point K.

2.540◉ Since line ZF is parallel to [line] QT [by construction], while [line] ZK is parallel to [line] OC [by construction], it is clear that plane ZKF will be parallel to plane OCT, which lies within plane AQT [since O is where HK intersects plane AQT]. Plane HFK intersects these two planes along lines CT and KF. Hence, CT and KF are parallel.

2.541◉ From point T draw [line] TP perpendicular to line ZF. Since it falls between two parallels [i.e., TQ and ZF], it is obvious that it will be parallel to line NZ, and so it will be tangent to the circle [at point T].⁑ Hence, plane ATP is tangent to the cone along line [of longitude] AT, and line OC is perpendicular to this plane. Accordingly, plane ATQ will be orthogonal to plane ATP, and plane ATP intersects the two planes ATQ and ZKF, which are parallel. Thus, the common sections, one being CT, the other PI, are parallel. But it has already been shown that CT is parallel to KF. Hence, PI is parallel to KF.

2.543◉ Now, if line CI is drawn, then, since plane ATPI is perpendicular to plane ZKF [which is parallel to plane ATQ], CI will be perpendicular to ZK, and angle CKZ = angle KZC. But angle ECO = angle CKZ [by Euclid, I.29, because OC and KZ are parallel, and EC cuts them both], and angle OCZ = [alternate] angle CZK, so angle ECO = angle OCZ. Hence, [the form of point] E is reflected to
[point] Z from point C, which is what was proposed.

2.544◉ Moreover, if another point is taken on the circle from which [the form of point] H is reflected to [point] Z, it will be demonstrated that [the form of point] E is reflected to Z from some point other than C on the cone. And if [the form of point] H is reflected to [point] Z from three points on the circle, [the form of point] E will be reflected to [point] Z from three [points] on
the cone; if [it is reflected] from four [points on the circle, it will be reflected] from four [points on the cone].

2.545◉ Furthermore, the point of reflection from which [the form of point] E is reflected to [point] Z is easy to find when the point on the circle from which [the form of] point H is reflected to [point] Z is found, and it will be found in the preceding way.

2.546◉ If, however, it were claimed that [the form of] point E could be reflected to [point] Z from more than four points on the cone, it would be possible by repeating the earlier proof to show that [the form of] point H is reflected to [point] Z from more than four points on the circle, and in the case where [the form of] point E will happen to be reflected to [point] Z from however many
points on the circle, or from one only, [the form of] point E will happen to be reflected to [point] Z from that many points on the cone, or from one only, and vice-versa. But if [this] contrary claim is made, it can be disproven in the preceding way [i.e., according to proposition 49].

2.547◉ Hence, it is clear that some of the points [that are reflected] have a single image, some [have] two, some three, and some four, but no [more] than four are possible. In addition, when the mirror is exposed to both eyes, the same image will have different locations, but, because of its imperceptibility, this difference [in location] does not cause an error [in visual perception].